NCERT Solutions · Class 12 · Chapter 10: Vector Algebra
NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2
Exercise 10.2: Components, unit vectors and the section formula. For \(\vec r = x\hat i + y\hat j + z\hat k\): \(|\vec r| = \sqrt{x^2 + y^2 + z^2}\), unit vector \(\hat r = \dfrac{\vec r}{|\vec r|}\), direction cosines \(\dfrac{x}{|\vec r|}, \dfrac{y}{|\vec r|}, \dfrac{z}{|\vec r|}\). \(\overrightarrow{PQ}\) = (position vector of Q) − (position vector of P). Section formula: \(\dfrac{m\vec b + n\vec a}{m + n}\) (internal), \(\dfrac{m\vec b - n\vec a}{m - n}\) (external).
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Exercise 10.2 questions and solutions
Exercise 10.2, Question 1
Find the magnitudes of \(\vec a = \hat i + \hat j + \hat k\), \(\vec b = 2\hat i - 7\hat j - 3\hat k\), \(\vec c = \tfrac{1}{\sqrt3}\hat i + \tfrac{1}{\sqrt3}\hat j - \tfrac{1}{\sqrt3}\hat k\).
Write two different vectors with the same magnitude.
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Any two different vectors whose components have the same sum of squares, e.g. \(\hat i + 2\hat j + 3\hat k\) and \(3\hat i + 2\hat j + \hat k\) (both \(\sqrt{14}\)).
Answer: e.g. \(\hat i + 2\hat j + 3\hat k\) and \(3\hat i + 2\hat j + \hat k\)
Show A, B, C with position vectors \(3\hat i - 4\hat j - 4\hat k\), \(2\hat i - \hat j + \hat k\), \(\hat i - 3\hat j - 5\hat k\) form a right-angled triangle.
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\[\overrightarrow{AB} = -\hat i + 3\hat j + 5\hat k\], \[\overrightarrow{BC} = -\hat i - 2\hat j - 6\hat k\], \[\overrightarrow{CA} = 2\hat i - \hat j + \hat k\]
\[\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\], so (A), (B) and (D) hold (in (D), \[-\overrightarrow{CB} = \overrightarrow{BC}\]).
(C) would need \[\overrightarrow{AC} = \overrightarrow{CA}\], false for a triangle. (The current reprint prints option (C) the same as (B); the option meant is \[\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{CA}\].)
\(\vec a\) and \(\vec b\) are collinear. Which are incorrect? (A) \(\vec b = \lambda\vec a\) for some scalar \(\lambda\) (B) \(\vec a = \pm\vec b\) (C) their components are not proportional (D) both have the same direction but different magnitudes
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Collinear means \(\vec b = \lambda\vec a\): (A) is correct, and it makes the components proportional, so (C) is incorrect.
(B) needs \(|\lambda| = 1\) and (D) needs \(\lambda > 0\), \(\lambda \ne 1\); neither holds for all collinear pairs.
Answer: (B), (C) and (D) are not true in general; only (A) always holds
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Vector Algebra has 39 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.
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.