These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-4/
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4
Class 12 Maths Ex 10.4 Question 1.
Find \(|\vec{a} \times \vec{b}|\), if \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\).
Solution:
Question 2.
Find a unit vector perpendicular to each of the vector \(\overrightarrow { a } +\overrightarrow { b } \quad and\quad \overrightarrow { a } -\overrightarrow { b } \), where \(\overrightarrow { a } =3\hat { i } +2\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =\hat { i } +2\hat { j } -2\hat { k } \)
Solution:
Question 3.
If a unit vector \(\vec{a}\) makes angles \(\frac { π }{ 3 }\) with \(\hat{i}\), \(\frac { π }{ 4 }\) with \(\hat{j}\) and an acute angle θ with k, then find θ and hence, the components of \(\vec{a}\).
Solution:
The direction cosines of \(\vec{a}\) are
Since \(\vec{a}\) is a unit vector, its components are the direction cosines
Question 4.
Show that
\((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Solution:
\((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b}\)
= \(\overrightarrow{0}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-\overrightarrow{0}\)
= 2(\(\vec{a}\) x \(\vec{b}\))
Question 5.
Find λ and μ if
\(\left( 2\hat { i } +6\hat { j } +27\hat { k } \right) \times \left( \hat { i } +\lambda \hat { j } +\mu \hat { k } \right)\) = \(\vec{0}\)
Solution:
\((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\)
\(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|=\overrightarrow{0}\)
\(\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}\)
Equating the corresponding components, we get
6µ – 27λ = 0 … (1)
– 2µ + 27 = 0 … (2)
2λ – 6 = 0 … (3)
(2) → µ = \(\frac { 27 }{ 2 }\), (3) → λ = 3
Substituting the values of λ and µ in (1),
we get 6(\(\frac { 27 }{ 2 }\)) – 27(3) = 81 – 81 = 0
(1) satisfy λ = 3 and µ = \(\frac { 27 }{ 2 }\)
Hence λ = 3, µ = \(\frac { 27 }{ 2 }\)
Question 6.
Given that \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = \(\vec{0}\) What can you conclude about the vectors \(\vec{a}\) and \(\vec{b}\)?
Solution:
\(\vec{a}\).\(\vec{b}\) = 0 ⇒ \(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\) = \(\vec{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\)
\(\vec{a}\) x \(\vec{b}\) = \(\vec{0}\) ⇒ \(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\) = \(\vec{0}\) or \(\vec{a}\) || \(\vec{b}\)
Since \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = 0,
then either \(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\).\(\vec{0}\)
\(\vec{a}\) ⊥ \(\vec{b}\) and \(\vec{a}\)||\(\vec{b}\) are not possible at the same time.
Question 7.
Let the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be given as \({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } ,{ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } ,{ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k }\), then show that \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\).
Solution:
Question 8.
If either \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text {, then } \vec{a} \times \vec{b}=\overrightarrow{0}\). Is the converse true? Justify your answer with an example.
Solution:
\(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}\)
If \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text {, then } \vec{a} \times \vec{b}=\overrightarrow{0}\).
The converse need not be true.
For example, consider the non-zero parallel
Question 9.
Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).
Solution:
Question 10.
Find the area of the parallelogram whose adjacent sides are determined by the vectors \(\overrightarrow { a } =\hat { i } -\hat { j } +3\hat { k } ,\overrightarrow { b } =2\hat { i } -7\hat { j } +\hat { k } \)
Solution:
Question 11.
Let the vectors \(\vec{a}\) and \(\vec{b}\) such that |\(\vec{a}\)| = 3 and |\(\vec{b}\)| = \(\frac{\sqrt{2}}{3}\), then \(\vec{a}\) x \(\vec{b}\) is a unit vector if the angle between \(\vec{a}\) and \(\vec{a}\) is
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 4 } \)
(c) \(\frac { \pi }{ 3 } \)
(d) \(\frac { \pi }{ 2 } \)
Solution:
Let θ be the angle between vectors \(\vec{a}\) and \(\vec{b}\).
Since \(\vec{a}\) x \(\vec{b}\) is a unit vector,
Question 12.
Area of a rectangles having vertices
\(\left( -\hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right), \left( \hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right)\)
\(\left( \hat { i } -\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right), \left( -\hat { i } -\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right)\)
(a) \(\frac { 1 }{ 2 }\) sq units
(b) 1 sq.units
(c) 2 sq.units
(d) 4 sq.units
Solution: