NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Question 1.
Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes \(\sqrt{3}\) and 2 respectively, and such that \(\vec{a}\).\(\vec{b}\) = \(\sqrt{6}\)
Solution:
|\(\vec{a}\)|= \(\sqrt{3}\)
and \(\vec{b}\) = 2, \(\vec{a}\).\(\vec{b}\) = \(\sqrt{6}\)
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{\sqrt{6}}{(\sqrt{3})(2)}=\frac{1}{\sqrt{2}}\)
∴ θ = \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)

Question 2.
Find the angle between the vectors \(\hat{i}-2 \hat{j}+3 \hat{k} \text { and } 3 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:
Let \(\vec{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}\) = \(\hat{3i}-2 \hat{j}+ \hat{k}\)
\(\vec{a}\).\(\vec{b}\) = \((\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})\)
= 1(3) + (- 2)(- 2) + 3(1) = 3 + 4 + 3 = 10
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 1

Question 3.
Find the projection of the vector \(\overrightarrow { i } -\overrightarrow { j }\), on the line represented by the vector \(\overrightarrow { i } +\overrightarrow { j }\).
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 2

Question 4.
Find the projection of the vector \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(7 \hat{i}-\hat{j}+8 \hat{k}\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 3

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Question 5.
Show that each of the given three vectors is a unit vector \(\frac { 1 }{ 7 } \left( 2\hat { i } +3\hat { j } +6\hat { k } \right) ,\frac { 1 }{ 7 } \left( 3\hat { i } -6\hat { j } +2\hat { k } \right) ,\frac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right)\)
Also show that they are mutually perpendicular to each other.
Solution:
Let \(\vec{a}\) = \(\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})\), \(\vec{b}\) = \(\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})\) and \(\vec{c}\) = \(\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\)
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 4
Here \(\vec{a}\).\(\vec{b}\) = 0, \(\vec{b}\).\(\vec{c}\) = 0 and \(\vec{a}\).\(\vec{c}\) = 0
∴ The vectors \(\vec{a}\).\(\vec{b}\) and \(\vec{c}\) are mutually per-pendicular vectors.

Question 6.
\(Find\left| \overrightarrow { a } \right| and\left| \overrightarrow { b } \right| if\left( \overrightarrow { a } +\overrightarrow { b } \right) \cdot \left( \overrightarrow { a } -\overrightarrow { b } \right) =8\quad and\left| \overrightarrow { a } \right| =8\left| \overrightarrow { b } \right| \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 5

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Question 7.
Evaluate the product :
\((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\)
Solution:
\(\left( 3\overrightarrow { a } -5\overrightarrow { b } \right) \cdot \left( 2\overrightarrow { a } +7\overrightarrow { b } \right) \)
\(=6\overrightarrow { a } .\overrightarrow { a } -10\overrightarrow { b } \overrightarrow { a } +21\overrightarrow { a } .\overrightarrow { b } -35\overrightarrow { b } .\overrightarrow { b } \)
\(=6{ \left| \overrightarrow { a } \right| }^{ 2 }-11\overrightarrow { a } \overrightarrow { b } -35{ \left| \overrightarrow { b } \right| }^{ 2 }\)

Question 8.
Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac { 1 }{ 2 }\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 6

Question 9.
Find |\(\vec{a}\)| , if for a unit vector \(\vec{a}\), \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 12
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 7

Question 10.
If \(\overrightarrow { a } =2\hat { i } +2\hat { j } +3\hat { k } ,\overrightarrow { b } =-\hat { i } +2\hat { j } +\hat { k } and \overrightarrow { c } =3\hat { i } +\hat { j } \) such that \(\overrightarrow { a } +\lambda \overrightarrow { b } \bot \overrightarrow { c } \) , then find the value of λ.
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 8

Question 11.
Show that \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a} \), is per-pendicular to \(|\vec{a}| \vec{b}-|\vec{b}| \vec{a} \) for any two non-zero vectors \(\vec{a} \text { and } \vec{b}\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 9

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Question 12.
If \(\overrightarrow { a } \cdot \overrightarrow { a } =0\quad and\quad \overrightarrow { a } \cdot \overrightarrow { b } =0\), then what can be concluded about the vector \(\overrightarrow { b } \) ?
Solution:
\(\vec{a}\).\(\vec{a}\) = 0 ⇒ \(\vec{a}\) is a zero vector.
since \(\vec{a}\) = \(\vec{0}\), \(\vec{a}\).\(\vec{b}\) = 0 for any vector \(\vec{b}\)
Vector \(\vec{b}\) be any vector.

Question 13.
If \(\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } \) are the unit vector such that \(\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0\) , then find the value of \(\overrightarrow { a } .\overrightarrow { b } +\overrightarrow { b } .\overrightarrow { c } +\overrightarrow { c } .\overrightarrow { a } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 10

Question 14.
If either vector \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}\), then \(\vec{a} \cdot \vec{b}\). But the converse need not be true. Justify your answer with an example.
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 11
Thus two non-zero vectors \(\vec{a}\) and \(\vec{b}\) may have \(\vec{a}\).\(\vec{a}\) zero.

Question 15.
If the vertices A,B,C of a triangle ABC are (1, 2, 3) (-1, 0, 0), (0, 1, 2) respectively, then find ∠ABC.
Solution:
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 12

Question 16.
Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are collinear.
Solution:
A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are the points.
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 13
∴ \(\overrightarrow{AB}\)\(\overrightarrow{AC}\) are parallel and A is a common point. Therefore A, B, C are collinear.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Question 17.
Show that the vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k} \text { and } \quad 3 \hat{i}-4 \hat{j}-4 \hat{k}\) from the vertices of a right angled triangle.
Solution:
Let A, B and C be the vertices of the triangle with the vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k} \text { and } \quad 3 \hat{i}-4 \hat{j}-4 \hat{k}\)
∴ \(\overrightarrow{AB}\) = p.v. of B – p.v. of A
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 14
∴ A, B, C are the vertices of ∆ ABC
\(\overrightarrow{BC}\).\(\overrightarrow{CA}\) = (2)(-1) + (-1)(3) + (1)(5)
= – 2 – 3 + 5 = 0
∴ \(\overrightarrow{BC}\)⊥\(\overrightarrow{CA}\)
Hence triangle ABC is a right triangle

Question 18.
If \(\overrightarrow { a } \) is a non-zero vector of magnitude ‘a’ and λ is a non- zero scalar, then λ \(\overrightarrow { a } \) is unit vector if
(a) λ = 1
(b) λ = – 1
(c) a = |λ|
(d) a = \(\frac { 1 }{ \left| \lambda \right| } \)
Solution:
\(\left| \overrightarrow { a } \right| =a\)
Given : \(\lambda \overrightarrow { a } \) is a unit vectors.
\(|\lambda \vec{a}|=1 \Rightarrow|\lambda||\vec{a}|=1 \Rightarrow|\lambda| a=1 \Rightarrow a=\frac{1}{|\lambda|}\)

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