NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 12 - Vector Algebra

NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercise gives explanations to 19 questions given in the Class 12 Mathematics NCERT book. Class 12 Maths chapter 10 miscellaneous exercise solutions cover questions from all the main topics of the chapter. Compared to the exercise questions NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises are a bit more tricky. Try to solve all these questions yourself before going to the Class 12 Maths chapter 10 miscellaneous solutions. So that students can understand their depth of understanding of the concepts and revisit or revise the concepts required. A total of five exercises including miscellaneous are present in the NCERT Class 12 Book. On this page, the NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises are given. Along with the NCERT questions students can practice NCERT exemplar and also Class 12 CBSE Previous Year Questions.

Vector Algebra Exercise 10.1
Vector Algebra Exercise 10.2
Vector Algebra Exercise 10.3
Vector Algebra Exercise 10.4

Vector Algebra Class 12 Chapter 10-Miscellaneous Exercise

Question:1 Write down a unit vector in XY-plane, making an angle of $30 \degree$ with the positive direction of x-axis.

Answer:

As we know

a unit vector in XY-Plane making an angle $\theta$ with x-axis :

$\vec r=cos\theta \hat i+sin\theta \hat j$

Hence for $\theta = 30^0$

$\vec r=cos(30^0) \hat i+sin(30^0) \hat j$

$\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j$

Answer- the unit vector in XY-plane, making an angle of $30 \degree$ with the positive direction of x-axis is

$\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j$

Question:2 Find the scalar components and magnitude of the vector joining the points
$P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).$

Answer:

Given in the question

$P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).$

And we need to finrd the scalar components and magnitude of the vector joining the points P and Q

$\vec {PQ}=(x_2-x_1)\hat i +(y_2-y_1)\hat j+(z_2-z_1)\hat k$

Magnitiude of vector PQ

$|\vec {PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Scalar components are

$(x_2-x_1),(y_2-y_1),(z_2-z_1)$

Question:3 A girl walks 4 km towards west, then she walks 3 km in a direction $30 \degree$ east of north and stops. Determine the girl’s displacement from her initial point of departure.

Answer:

As the girl walks 4km towards west

Position vector = $-4\hat i$

Now as she moves 3km in direction 30 degree east of north.

$-4\hat i+3sin30^0\hat i+3cos30^0\hat j$

$-4\hat i+\frac{3}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

$\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

hence final position vector is;

$\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

Question:4 If $\vec a = \vec b + \vec c$ , then is it true that $|\vec a| =| \vec b |+| \vec c |$ ? Justify your answer.

Answer:

No, if $\vec a = \vec b + \vec c$ then we can not conclude that $|\vec a| =| \vec b |+| \vec c |$ .

the condition $\vec a = \vec b + \vec c$ satisfies in the triangle.

also, in a triangle, $|\vec a| <| \vec b |+| \vec c |$

Since, the condition $|\vec a| =| \vec b |+| \vec c |$ is contradicting with the triangle inequality, if $\vec a = \vec b + \vec c$ then we can not conclude that $|\vec a| =| \vec b |+| \vec c |$

Question:5 Find the value of x for which $x ( \hat i+ \hat j + \hat k )$ is a unit vector.

Answer:

Given in the question,

a unit vector, $\vec u=x ( \hat i+ \hat j + \hat k )$

We need to find the value of x

$|\vec u|=1$

$|x ( \hat i+ \hat j + \hat k )|=1$

$x\sqrt{1^2+1^2+1^2}=1$

$x\sqrt{3}=1$

$x=\frac{1}{\sqrt{3}}$

The value of x is $\frac{1}{\sqrt{3}}$

Question:6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec a = 2 \hat i + 3 \hat j - \hat k \: \: and \: \: \vec b = \hat i - 2 \hat j + \hat k$

Answer:

Given two vectors

$\vec a = 2 \hat i + 3 \hat j - \hat k \: \: and \: \: \vec b = \hat i - 2 \hat j + \hat k$

Resultant of $\vec a$ and $\vec b$ :

$\vec R = \vec a +\vec b$ $=2 \hat i + 3 \hat j - \hat k + \hat i - 2 \hat j + \hat k=3\hat i + \hat j$

Now, a unit vector in the direction of $\vec R$

$\vec u =\frac{3\hat i+\hat j}{\sqrt{3^2+1^2}}=\frac{3}{\sqrt{10}}\hat i+\frac{1}{\sqrt{10}}\hat j$

Now, a unit vector of magnitude in direction of $\vec R$

$\vec v=5\vec u =5\frac{3}{\sqrt{10}}\hat i+5\frac{1}{\sqrt{10}}\hat j=\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

Hence the required vector is $\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

Question:7 If $\vec a = \hat i + \hat j + \hat k , \vec b = 2 \hat i - \hat j + 3 \hat k \: \: and\: \: \vec c = \hat i - 2 \hat j + \hat k$ , find a unit vector parallel to the vector $2\vec a - \vec b + 3 \vec c$ .

Answer:

Given in the question,

$\vec a = \hat i + \hat j + \hat k , \vec b = 2 \hat i - \hat j + 3 \hat k \: \: and\: \: \vec c = \hat i - 2 \hat j + \hat k$

Now,

let vector $\vec V=2\vec a - \vec b + 3 \vec c$

$\vec V=2(\hat i +\hat j +\hat k) - (2\hat i-\hat j+3\hat k)+ 3 (\hat i-2\hat j+\hat k)$

$\vec V=3\hat i-3\hat j+2\hat k$

Now, a unit vector in direction of $\vec V$

$\vec u =\frac{3\hat i-3\hat j+2\hat k}{\sqrt{3^2+(-3)^2+2^2}}=\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k$

Now,

A unit vector parallel to $\vec V$

$\vec u =\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k$

OR

$-\vec u =-\frac{3}{\sqrt{22}}\hat i+\frac{3}{\sqrt{22}} \hat j-\frac{2}{\sqrt{22}}\hat k$

Question:8 Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

$\vec {AB }=(5-1)\hat i+(0-(-2))\hat j+(-2-(-8))\hat k=4\hat i+2\hat j+6\hat k$

$\vec {BC }=(11-5)\hat i+(3-0)\hat j+(7-(-2))\hat k=6\hat i+3\hat j+9\hat k$

$\vec {CA }=(11-1)\hat i+(3-(-2))\hat j+(7-(-8))\hat k=10\hat i+5\hat j+15\hat k$

now let's calculate the magnitude of the vectors

$|\vec {AB }|=\sqrt{4^2+2^2+6^2}=\sqrt{56}=2\sqrt{14}$

$|\vec {BC }|=\sqrt{6^2+3^2+9^2}=\sqrt{126}=3\sqrt{14}$

$|\vec {CA }|=\sqrt{10^2+5^2+15^2}=\sqrt{350}=5\sqrt{14}$

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio 2 : 3.

Question:9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $( 2 \vec a + \vec b ) \: \:and \: \: ( \vec a - 3 \vec b )$ externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Answer:

Given, two vectors $\vec P=( 2 \vec a + \vec b ) \: \:and \: \:\vec Q= ( \vec a - 3 \vec b )$

the point R which divides line segment PQ in ratio 1:2 is given by

$=\frac{2(2\vec a +\vec b)-(\vec a-3\vec b)}{2-1}=4\vec a +2\vec b -\vec a+3\vec b=3\vec a+5\vec b$

Hence position vector of R is $3\vec a+5\vec b$ .

Now, Position vector of the midpoint of RQ

$=\frac{( 3\vec a + 5\vec b + \vec a - 3 \vec b )}{2}=2\vec a+\vec b$

which is the position vector of Point P . Hence, P is the mid-point of RQ

Question:10 The two adjacent sides of a parallelogram are $2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k$ . Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

Given, two adjacent sides of the parallelogram

$2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k$

The diagonal will be the resultant of these two vectors. so

resultant R:

$\vec R=2 \hat i - 4 \hat j + 5 \hat k \: +\: \hat i - 2 \hat j - 3 \hat k=3\hat i-6\hat j+2\hat k$

Now unit vector in direction of R

$\vec u=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{49}}=\frac{3\hat i-6\hat j+2\hat k}{7}$

Hence unit vector along the diagonal of the parallelogram

$\vec u={\frac{3}{7}\hat i-\frac{6}{7}\hat j+\frac{2}{7}\hat k}$

Now,

Area of parallelogram

$A=(2 \hat i - 4 \hat j + 5 \hat k )\: \times \: \: (\hat i - 2 \hat j - 3 \hat k)$

$A=\begin{vmatrix} \hat i &\hat j &\hat k \\ 2& -4 &5 \\ 1&-2 &-3 \end{vmatrix}=|\hat i(12+10)-\hat j(-6-5)+\hat k(-4+4)|=|22\hat i +11\hat j|$

$A=\sqrt{22^2+11^2}=11\sqrt{5}$

Hence the area of the parallelogram is $11\sqrt{5}$ .

Question:11 Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm \left ( \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } \right )$

Answer:

Let a vector $\vec a$ is equally inclined to axis OX, OY and OZ.

let direction cosines of this vector be

$cos\alpha,cos\alpha \:and \:cos\alpha$

Now

$cos^2\alpha+cos^2\alpha +cos^2\alpha=1$

$cos^2\alpha=\frac{1}{3}$

$cos\alpha=\frac{1}{\sqrt{3}}$

Hence direction cosines are:

$\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )$

Question:12 Let $\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k$ . Find a vector $\vec d$ which is perpendicular to both $\vec a \: \: and \: \: \vec b \: \: and \: \: \vec c . \vec d = 15$

Answer:

Given,

$\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k$

Let $\vec d=d_1\hat i+d_2\hat j +d_3\hat k$

now, since it is given that d is perpendicular to $\vec a$ and $\vec b$ , we got the condition,

$\vec b.\vec d=0$ and $\vec a.\vec d=0$

$(\hat i+4\hat j +2\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$ And $(3\hat i-2\hat j +7\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$

$d_1+4d_2+2d_3=0$ And $3d_1-2d_2+7d_3=0$

here we got 2 equation and 3 variable. one more equation will come from the condition:

$\vec c . \vec d = 15$

$(2\hat i-\hat j +4\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=15$

$2d_1-d_2+4d_3=15$

so now we have three equation and three variable,

$d_1+4d_2+2d_3=0$

$3d_1-2d_2+7d_3=0$

$2d_1-d_2+4d_3=15$

On solving this three equation we get,

$d_1=\frac{160}{3},d_2=-\frac{5}{3}\:and\:d_3=-\frac{70}{3}$ ,

Hence Required vector :

$\vec d=\frac{160}{3}\hat i-\frac{5}{3}\hat j-\frac{70}{3}\hat k$

Question:13 The scalar product of the vector $\hat i + \hat j + \hat k$ with a unit vector along the sum of vectors $2\hat i + 4 \hat j -5 \hat k$ and $\lambda \hat i + 2 \hat j +3 \hat k$ is equal to one. Find the value of $\lambda$ .

Answer:

Let, the sum of vectors $2\hat i + 4 \hat j -5 \hat k$ and $\lambda \hat i + 2 \hat j +3 \hat k$ be

$\vec a=(\lambda +2)\hat i + 6 \hat j -2 \hat k$

unit vector along $\vec a$

$\vec u=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{(\lambda+2)^2+6^2+(-2)^2}}=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}$

Now, the scalar product of this with $\hat i + \hat j + \hat k$

$\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}.(\hat i+\hat j +\hat k)$

$\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\frac{(\lambda +6) }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\lambda =1$

Question:14 If $\vec a , \vec b , \vec c$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec a+\vec b +\vec c$ is equally inclined to $\vec a , \vec b \: \: and \: \: \vec c$ .

Answer:

Given

$|\vec a|=|\vec b|=|\vec c|$ and

$\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=0$

Now, let vector $\vec a+\vec b +\vec c$ is inclined to $\vec a , \vec b \: \: and \: \: \vec c$ at $\theta_1,\theta_2\:and\:\theta_3$ respectively.

Now,

$cos\theta_1=\frac{(\vec a+\vec b+\vec c).\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a +\vec a.\vec b +\vec c.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{|\vec a|}{|\vec a+\vec b+\vec c|}$

$cos\theta_2=\frac{(\vec a+\vec b+\vec c).\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec a.\vec b +\vec b.\vec b +\vec c.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec b.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{|\vec b|}{|\vec a+\vec b+\vec c|}$

$cos\theta_3=\frac{(\vec a+\vec b+\vec c).\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec a.\vec c +\vec b.\vec c +\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{|\vec c|}{|\vec a+\vec b+\vec c|}$

Now, Since, $|\vec a|=|\vec b|=|\vec c|$

$cos\theta_1=cos\theta_2=cos\theta_3$

$\theta_1=\theta_2=\theta_3$

Hence vector $\vec a+\vec b +\vec c$ is equally inclined to $\vec a , \vec b \: \: and \: \: \vec c$ .

Question:15 Prove that $( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2$ , if and only if $\vec a , \vec b$ are perpendicular, given $\vec a \neq 0 , \vec b \neq 0$

Answer:

Given in the question,

$\vec a , \vec b$ are perpendicular and we need to prove that $( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2$

LHS= $( \vec a + \vec b ) . (\vec a + \vec b ) = \vec a .\vec a+\vec a.\vec b+\vec b.\vec a+\vec b.\vec b$

$= \vec a .\vec a+2\vec a.\vec b+\+\vec b.\vec b$

$= |\vec a |^2+2\vec a.\vec b+\+|\vec b|^2$

if $\vec a , \vec b$ are perpendicular, $\vec a.\vec b=0$

$( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a |^2+2\vec a.\vec b+\+|\vec b|^2$

$= |\vec a |^2+|\vec b|^2$

= RHS

LHS ie equal to RHS

Hence proved.

Question:16 Choose the correct answer If $\theta$ is the angle between two vectors $\vec a \: \: and \: \: \vec b$ , then $\vec a \cdot \vec b \geq 0$ only when
$\\A ) 0 < \theta < \frac{\pi }{2} \\\\ \: \: \: \: B ) 0 \leq \theta \leq \frac{\pi }{2} \\\\ \: \: \: C ) 0 < \theta < \pi \\\\ \: \: \: D) 0 \leq \theta \leq\pi$

Answer:

Given in the question

$\theta$ is the angle between two vectors $\vec a \: \: and \: \: \vec b$

$\vec a \cdot \vec b \geq 0$

$|\vec a| | \vec b |cos\theta\geq 0$

this will satisfy when

$cos\theta\geq 0$

$0\leq\theta\leq \frac{\pi}{2}$

Hence option B is the correct answer.

Question:17 Choose the correct answer. Let $\vec a \: \: and \: \: \vec b$ be two unit vectors and $\theta$ is the angle between them. Then $\vec a + \vec b$ is a unit vector if

$\\A ) \theta = \frac{\pi }{4} \\\\ B ) \theta = \frac{\pi }{3} \\\\ C ) \theta = \frac{\pi }{2} \\\\ D ) \theta = \frac{2\pi }{3}$

Answer:

Gicen in the question

$\vec a \: \: and \: \: \vec b$ be two unit vectors and $\theta$ is the angle between them

$|\vec a|=1,\:and\:\:|\vec b|=1$

also

$|\vec a + \vec b|=1$

$|\vec a + \vec b|^2=1$

$|\vec a|^2 + |\vec b|^2+2\vec a.\vec b=1$

$1 + 1+2\vec a.\vec b=1$

$\vec a.\vec b=-\frac{1}{2}$

$|\vec a||\vec b|cos\theta =-\frac{1}{2}$

$cos\theta =-\frac{1}{2}$

$\theta =\frac{2\pi}{3}$

Then $\vec a + \vec b$ is a unit vector if $\theta =\frac{2\pi}{3}$

Hence option D is correct.

Question:18 The value of $\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j )$ is

(A) 0

(B) –1

(C) 1

(D) 3

Answer:

To find the value of $\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j )$

$\\\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j ) \\=\hat i.\hat i+\hat j(-\hat j)+\hat k.\hat k\\=1-1+1\\=1$

Hence option C is correct.

Question:19 Choose the correct. If $\theta$ is the angle between any two vectors $\vec a \: \:and \: \: \vec b$ , then $|\vec a \cdot \vec b |=|\vec a \times \vec b |$ when $\theta$
is equal to

$\\A ) 0 \\\\ B ) \pi /4 \\\\ C ) \pi / 2 \\\\ D ) \pi$

Answer:

Given in the question

$\theta$ is the angle between any two vectors $\vec a \: \:and \: \: \vec b$ and $|\vec a \cdot \vec b |=|\vec a \times \vec b |$

To find the value of $\theta$

Hence option D is correct.

Benefits of ncert solutions for Class 12 Maths chapter 10 miscellaneous exercises

By using miscellaneous exercise chapter 10 Class 12 students will be able to get an idea of the complete chapter.
Each question explained in the Class 12 Maths chapter 10 miscellaneous solutions are useful for the CBSE Class 12 board exam and also for exams like JEE main, VITEEE, BITSAT, etc.
The students can revise the chapter vector algebra by practising the ncert solutions for Class 12 Maths chapter 10 miscellaneous exercises

Also see-

NCERT Exemplar Solutions Class 12 Maths Chapter 10
NCERT Solutions for Class 12 Maths Chapter 10

NCERT Solutions Subject Wise

NCERT Solutions Class 12 Chemistry
NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Biology
NCERT Solutions for Class 12 Mathematics

Subject Wise NCERT Exemplar Solutions

NCERT Exemplar Class 12 Maths
NCERT Exemplar Class 12 Physics
NCERT Exemplar Class 12 Chemistry
NCERT Exemplar Class 12 Biology

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 12 - Vector Algebra

Vector Algebra Class 12 Chapter 10-Miscellaneous Exercise

Question:1 Write down a unit vector in XY-plane, making an angle of with the positive direction of x-axis.

Answer:

As we know

a unit vector in XY-Plane making an angle with x-axis :

Hence for

Answer- the unit vector in XY-plane, making an angle of with the positive direction of x-axis is

Question:2 Find the scalar components and magnitude of the vector joining the points

Answer:

Given in the question

And we need to finrd the scalar components and magnitude of the vector joining the points P and Q

Magnitiude of vector PQ

Scalar components are

Question:3 A girl walks 4 km towards west, then she walks 3 km in a direction east of north and stops. Determine the girl’s displacement from her initial point of departure.

Answer:

As the girl walks 4km towards west

Position vector =

Now as she moves 3km in direction 30 degree east of north.

hence final position vector is;

Question:4 If , then is it true that ? Justify your answer.

Answer:

No, if then we can not conclude that .

the condition satisfies in the triangle.

also, in a triangle,

Since, the condition is contradicting with the triangle inequality, if then we can not conclude that

Question:5 Find the value of x for which is a unit vector.

Answer:

Given in the question,

a unit vector,

We need to find the value of x

The value of x is

Question:6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

Answer:

Given two vectors

Resultant of and :

Now, a unit vector in the direction of

Now, a unit vector of magnitude in direction of

Hence the required vector is

Question:7 If , find a unit vector parallel to the vector .

Answer:

Given in the question,

Now,

let vector

Now, a unit vector in direction of

Now,

A unit vector parallel to

OR

Question:8 Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

now let's calculate the magnitude of the vectors

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio 2 : 3.

Question:9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Answer:

Given, two vectors

the point R which divides line segment PQ in ratio 1:2 is given by

Hence position vector of R is .

Now, Position vector of the midpoint of RQ

which is the position vector of Point P . Hence, P is the mid-point of RQ

Question:10 The two adjacent sides of a parallelogram are . Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

Given, two adjacent sides of the parallelogram

The diagonal will be the resultant of these two vectors. so

resultant R:

Now unit vector in direction of R

Hence unit vector along the diagonal of the parallelogram

Now,

Area of parallelogram

Hence the area of the parallelogram is .

Question:11 Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

Answer:

Let a vector is equally inclined to axis OX, OY and OZ.

let direction cosines of this vector be

Now

Hence direction cosines are:

Question:12 Let . Find a vector which is perpendicular to both

Question:1 Write down a unit vector in XY-plane, making an angle of $30 \degree$ with the positive direction of x-axis.

a unit vector in XY-Plane making an angle $\theta$ with x-axis :

Hence for $\theta = 30^0$

Answer- the unit vector in XY-plane, making an angle of $30 \degree$ with the positive direction of x-axis is

Question:2 Find the scalar components and magnitude of the vector joining the points
$P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).$

Question:3 A girl walks 4 km towards west, then she walks 3 km in a direction $30 \degree$ east of north and stops. Determine the girl’s displacement from her initial point of departure.

Position vector = $-4\hat i$

Question:4 If $\vec a = \vec b + \vec c$ , then is it true that $|\vec a| =| \vec b |+| \vec c |$ ? Justify your answer.

No, if $\vec a = \vec b + \vec c$ then we can not conclude that $|\vec a| =| \vec b |+| \vec c |$ .

the condition $\vec a = \vec b + \vec c$ satisfies in the triangle.

also, in a triangle, $|\vec a| <| \vec b |+| \vec c |$

Since, the condition $|\vec a| =| \vec b |+| \vec c |$ is contradicting with the triangle inequality, if $\vec a = \vec b + \vec c$ then we can not conclude that $|\vec a| =| \vec b |+| \vec c |$

Question:5 Find the value of x for which $x ( \hat i+ \hat j + \hat k )$ is a unit vector.

a unit vector, $\vec u=x ( \hat i+ \hat j + \hat k )$

The value of x is $\frac{1}{\sqrt{3}}$

Question:6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec a = 2 \hat i + 3 \hat j - \hat k \: \: and \: \: \vec b = \hat i - 2 \hat j + \hat k$

Resultant of $\vec a$ and $\vec b$ :

Now, a unit vector in the direction of $\vec R$

Now, a unit vector of magnitude in direction of $\vec R$

Hence the required vector is $\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

Question:7 If $\vec a = \hat i + \hat j + \hat k , \vec b = 2 \hat i - \hat j + 3 \hat k \: \: and\: \: \vec c = \hat i - 2 \hat j + \hat k$ , find a unit vector parallel to the vector $2\vec a - \vec b + 3 \vec c$ .

let vector $\vec V=2\vec a - \vec b + 3 \vec c$

Now, a unit vector in direction of $\vec V$

A unit vector parallel to $\vec V$

Question:9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $( 2 \vec a + \vec b ) \: \:and \: \: ( \vec a - 3 \vec b )$ externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Given, two vectors $\vec P=( 2 \vec a + \vec b ) \: \:and \: \:\vec Q= ( \vec a - 3 \vec b )$

Hence position vector of R is $3\vec a+5\vec b$ .

Question:10 The two adjacent sides of a parallelogram are $2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k$ . Find the unit vector parallel to its diagonal. Also, find its area.

Hence the area of the parallelogram is $11\sqrt{5}$ .

Question:11 Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm \left ( \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } \right )$

Let a vector $\vec a$ is equally inclined to axis OX, OY and OZ.

Question:12 Let $\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k$ . Find a vector $\vec d$ which is perpendicular to both $\vec a \: \: and \: \: \vec b \: \: and \: \: \vec c . \vec d = 15$

Let $\vec d=d_1\hat i+d_2\hat j +d_3\hat k$

now, since it is given that d is perpendicular to $\vec a$ and $\vec b$ , we got the condition,

$\vec b.\vec d=0$ and $\vec a.\vec d=0$

$(\hat i+4\hat j +2\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$ And $(3\hat i-2\hat j +7\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$

$d_1+4d_2+2d_3=0$ And $3d_1-2d_2+7d_3=0$

$d_1=\frac{160}{3},d_2=-\frac{5}{3}\:and\:d_3=-\frac{70}{3}$ ,

Question:13 The scalar product of the vector $\hat i + \hat j + \hat k$ with a unit vector along the sum of vectors $2\hat i + 4 \hat j -5 \hat k$ and $\lambda \hat i + 2 \hat j +3 \hat k$ is equal to one. Find the value of $\lambda$ .

Let, the sum of vectors $2\hat i + 4 \hat j -5 \hat k$ and $\lambda \hat i + 2 \hat j +3 \hat k$ be