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. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-3/
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3
Class 12 Maths Ncert Solutions Chapter 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.
\(|\lambda \vec{a}|=1 \Rightarrow|\lambda||\vec{a}|=1 \Rightarrow|\lambda| a=1 \Rightarrow a=\frac{1}{|\lambda|}\)