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NCERT Solutions · Class 12 · Chapter 10: Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4

Exercise 10.4: Vector (cross) product. \(\vec a \times \vec b = \begin{vmatrix}\hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix}\), perpendicular to both, with \(|\vec a \times \vec b| = |\vec a||\vec b|\sin\theta\). Parallel (collinear) vectors have zero cross product. Area of parallelogram \(= |\vec a \times \vec b|\); of triangle ABC \(= \tfrac12|\overrightarrow{AB} \times \overrightarrow{AC}|\).

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Exercise 10.4 questions and solutions

Exercise 10.4, Question 1

Find \(|\vec a \times \vec b|\) for \(\vec a = \hat i - 7\hat j + 7\hat k\), \(\vec b = 3\hat i - 2\hat j + 2\hat k\).
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  1. \[\begin{aligned}\vec a \times \vec b &= \hat i(-14 + 14) - \hat j(2 - 21) + \hat k(-2 + 21) \\ &= 19\hat j + 19\hat k\end{aligned}\]
Answer: \(19\sqrt2\)

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Exercise 10.4, Question 2

Find a unit vector perpendicular to both \(\vec a + \vec b\) and \(\vec a - \vec b\), where \(\vec a = 3\hat i + 2\hat j + 2\hat k\), \(\vec b = \hat i + 2\hat j - 2\hat k\).
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  1. \(\vec a + \vec b = 4\hat i + 4\hat j\), \(\vec a - \vec b = 2\hat i + 4\hat k\).
  2. Cross product: \(16\hat i - 16\hat j - 8\hat k\), magnitude 24.
Answer: \(\pm\tfrac13(2\hat i - 2\hat j - \hat k)\)

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Exercise 10.4, Question 3

A unit vector \(\vec a\) makes angles \(\tfrac{\pi}{3}\) with \(\hat i\), \(\tfrac{\pi}{4}\) with \(\hat j\) and an acute angle \(\theta\) with \(\hat k\). Find \(\theta\) and the components of \(\vec a\).
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  1. \[\begin{aligned}&\cos^2\tfrac{\pi}{3} + \cos^2\tfrac{\pi}{4} + \cos^2\theta = 1 \\ \Rightarrow\ &\tfrac14 + \tfrac12 + \cos^2\theta = 1\end{aligned}\], so \(\cos\theta = \tfrac12\) (acute).
Answer: \(\theta = \tfrac{\pi}{3}\); \[\vec a = \tfrac12\hat i + \tfrac{1}{\sqrt2}\hat j + \tfrac12\hat k\]

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Exercise 10.4, Question 4

Show \((\vec a - \vec b) \times (\vec a + \vec b) = 2(\vec a \times \vec b)\).
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  1. Expand: \[\vec a \times \vec a + \vec a \times \vec b - \vec b \times \vec a - \vec b \times \vec b = \vec 0 + \vec a \times \vec b + \vec a \times \vec b - \vec 0\]
Answer: Shown

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Exercise 10.4, Question 5

Find \(\lambda\) and \(\mu\) if \((2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda\hat j + \mu\hat k) = \vec 0\).
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  1. Zero cross product means parallel: \[\begin{aligned}\dfrac12 &= \dfrac{\lambda}{6} \\ &= \dfrac{\mu}{27}\end{aligned}\]
Answer: \(\lambda = 3,\ \mu = \tfrac{27}{2}\)

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Exercise 10.4, Question 6

\(\vec a \cdot \vec b = 0\) and \(\vec a \times \vec b = \vec 0\). What can you conclude?
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  1. Non-zero vectors cannot be both perpendicular (\(\cos\theta = 0\)) and parallel (\(\sin\theta = 0\)).
Answer: \(\vec a = \vec 0\) or \(\vec b = \vec 0\)

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Exercise 10.4, Question 7

Show \(\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c\) for \(\vec a = a_1\hat i + a_2\hat j + a_3\hat k\), and similar \(\vec b, \vec c\).
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  1. Each component of \(\vec a \times (\vec b + \vec c)\) is a \(2 \times 2\) determinant with a column \(b_j + c_j\); splitting that column splits the determinant into the matching components of \(\vec a \times \vec b\) and \(\vec a \times \vec c\).
Answer: Shown

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Exercise 10.4, Question 8

If \(\vec a = \vec 0\) or \(\vec b = \vec 0\), then \(\vec a \times \vec b = \vec 0\). Is the converse true?
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  1. No: parallel non-zero vectors also give \(\vec 0\), e.g. \(\vec a = \hat i + \hat j\), \(\vec b = 2\hat i + 2\hat j\).
Answer: No (example: \[(\hat i + \hat j) \times (2\hat i + 2\hat j) = \vec 0\])

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Exercise 10.4, Question 9

Find the area of the triangle with vertices A\((1, 1, 2)\), B\((2, 3, 5)\), C\((1, 5, 5)\).
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  1. \[\overrightarrow{AB} = \hat i + 2\hat j + 3\hat k\], \(\overrightarrow{AC} = 4\hat j + 3\hat k\); \[\overrightarrow{AB} \times \overrightarrow{AC} = -6\hat i - 3\hat j + 4\hat k\]
Answer: \(\tfrac12\sqrt{61}\) square units

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Exercise 10.4, Question 10

Find the area of the parallelogram with adjacent sides \(\hat i - \hat j + 3\hat k\) and \(2\hat i - 7\hat j + \hat k\).
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  1. Cross product: \[\hat i(-1 + 21) - \hat j(1 - 6) + \hat k(-7 + 2) = 20\hat i + 5\hat j - 5\hat k\]
Answer: \(15\sqrt2\) square units

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Exercise 10.4, Question 11

\(|\vec a| = 3\), \(|\vec b| = \tfrac{\sqrt2}{3}\). \(\vec a \times \vec b\) is a unit vector if the angle is: (A) \(\tfrac{\pi}{6}\) (B) \(\tfrac{\pi}{4}\) (C) \(\tfrac{\pi}{3}\) (D) \(\tfrac{\pi}{2}\)
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  1. \[\begin{aligned}&3 \cdot \tfrac{\sqrt2}{3}\sin\theta = 1 \\ \Rightarrow\ &\sin\theta = \tfrac{1}{\sqrt2}\end{aligned}\]
Answer: (B)

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Exercise 10.4, Question 12

The area of the rectangle with vertices \(-\hat i + \tfrac12\hat j + 4\hat k\), \(\hat i + \tfrac12\hat j + 4\hat k\), \(\hat i - \tfrac12\hat j + 4\hat k\), \(-\hat i - \tfrac12\hat j + 4\hat k\) is: (A) \(\tfrac12\) (B) 1 (C) 2 (D) 4
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  1. \(\overrightarrow{AB} = 2\hat i\), \(\overrightarrow{BC} = -\hat j\): sides 2 and 1.
Answer: (C) 2

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Vector Algebra in our sample papers: Sample paper 1 (questions 13, 15, 24) · Sample paper 2 (questions 16, 19, 34) · Sample paper 3 (questions 13, 14, 24) · Sample paper 4 (questions 16, 19, 25) · Sample paper 5 (questions 13, 19, 24).

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Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.