First slide

Algebra of vectors

Question

If $\hat{i} - 3 \hat{j} + 5 \hat{k}$ bisects the angle between $\hat{a}$ and $- \hat{i} + 2 \hat{j} + 2 \hat{k},$ where $\hat{a}$ is a unit vector, then

Moderate

A

$\hat{a} = \frac{1}{105} (41 \hat{i} + 88 \hat{j} - 40 \hat{k})$
B

$\hat{a} = \frac{1}{105} (41 \hat{i} + 88 \hat{j} + 40 \hat{k})$
C

$\hat{a} = \frac{1}{105} (- 41 \hat{i} + 88 \hat{j} - 40 \hat{k})$
D

$\hat{a} = \frac{1}{105} (41 \hat{i} - 88 \hat{j} - 40 \hat{k})$

Solution

$We must have λ (\hat{i} - 3 \hat{j} + 5 \hat{k}) = \hat{a} + \frac{2 \hat{k} + 2 \hat{j} - \hat{i}}{3}$
Therefore,
$\begin{aligned} 3 \hat{a} = 3 λ (\hat{i} - 3 \hat{j} + 5 \hat{k}) - (2 \hat{k} + 2 \hat{j} - \hat{i}) \\ = \hat{i} (3 λ + 1) - \hat{j} (2 + 9 λ) + \hat{k} (15 λ - 2) \end{aligned}$
$\begin{array}{l} or 3 | \hat{a} | = \sqrt{(3 λ + 1)^{2} + (2 + 9 λ)^{2} + (15 λ - 2)^{2}} \\ or 9 = (3 λ + 1)^{2} + (2 + 9 λ)^{2} + (15 λ - 2)^{2} \\ or 315 λ^{2} - 18 λ = 0 \Rightarrow λ = 0, \frac{2}{35} \end{array}$
$If λ = 0, \vec{a} = \hat{i} - 2 \hat{j} - 2 \hat{k} (not acceptable)$
$For λ = \frac{2}{35}, \vec{a} = \frac{41}{105} \hat{i} - \frac{88}{105} \hat{j} - \frac{40}{105} \hat{k}$

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