(B) Given that $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$.Since $a$ and $b$ are collinear, $b = \lambda a$ for some scalar $\lambda$.We are given $|b| =

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$\pm 3(2 \hat{i} + 3 \hat{j} + 6 \hat{k})$

(B) Given that $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$.Since $a$ and $b$ are collinear, $b = \lambda a$ for some scalar $\lambda$.We are given $|b| = 21$.First, calculate the magnitude of $a$:$|a| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.Since $b = \lambda a$, we have $|b| = |\lambda| |a|$.$21 = |\lambda| \times 7 \implies |\lambda| = 3 \implies \lambda = \pm 3$.Therefore, $b = \pm 3(2 \hat{i} + 3 \hat{j} + 6 \hat{k})$.

Class 12 Mathematics Vector Algebra Basic , Modulus and Algebra of vectors

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If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$, then $b$ is equal to:

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A
$\pm(2 \hat{i} + 3 \hat{j} + 6 \hat{k})$
B
$\pm 3(2 \hat{i} + 3 \hat{j} + 6 \hat{k})$
C
$(\hat{i} + \hat{j} + \hat{k})$
D
$\pm 21(2 \hat{i} + 3 \hat{j} + 6 \hat{k})$

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If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$, then $b$ is equal to:

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