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.
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 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:
