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
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
Question 3.
Find the projection of the vector \(\overrightarrow { i } -\overrightarrow { j }\), on the line represented by the vector \(\overrightarrow { i } +\overrightarrow { j }\).
Solution:
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:
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 \(\vec{c}\) = \(\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\)
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:
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:
Question 9.
Find |\(\vec{a}\)| , if for a unit vector \(\vec{a}\), \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 12
Solution:
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:
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:
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:
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:
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:
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.
∴ \(\overrightarrow{AB}\)\(\overrightarrow{AC}\) are parallel and A is a common point. Therefore A, B, C are collinear.
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
∴ 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.
\(|\lambda \vec{a}|=1 \Rightarrow|\lambda||\vec{a}|=1 \Rightarrow|\lambda| a=1 \Rightarrow a=\frac{1}{|\lambda|}\)