First slide

Application of vectors in geometry

Question

A vector $\vec{a} = α \hat{i} + 2 \hat{j} + β \hat{k}, (α, β \in R)$ lies in the plane of the vectors $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{i} - \hat{j} + 4 \hat{k}$ . If $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$ , . then

Moderate

A

$\vec{a} . \hat{i} + 1 = 0$
B

$\vec{a} . \hat{i} + 3 = 0$
C

$\vec{a} . \hat{k} + 4 = 0$
D

$\vec{a} . \hat{k} + 2 = 0$

Solution

Angle bisector can be $\vec{a} = λ (\hat{b} + \hat{c}) o r \vec{a} = μ (\hat{b} - \hat{c})$
Now $\vec{a} = λ (\frac{\hat{i} + \hat{j}}{\sqrt{2}} + \frac{\hat{i} - \hat{j} + 4 \hat{k}}{3 \sqrt{2}}) = \frac{λ}{3 \sqrt{2}} [3 \hat{i} + 3 \hat{j} + \hat{i} - \hat{j} + 4 \hat{k}] = \frac{λ}{3 \sqrt{2}} [4 \hat{i} + 2 \hat{j} + 4 \hat{k}]$
Comparing with $\vec{a} = α \hat{i} + 2 \hat{j} + β \hat{k}$ we get
$\frac{2 λ}{3 \sqrt{2}} = 2 \Rightarrow λ = 3 \sqrt{2}$
$\vec{\Rightarrow a} = 4 \hat{i} + 2 \hat{j} + 4 \hat{k}$
          None of the option is satisfied. So, now consider
$\vec{a} = μ (\frac{\hat{i} + \hat{j}}{\sqrt{2}} - \frac{\hat{i} - \hat{j} + 4 \hat{k}}{3 \sqrt{2}})$
       $\Rightarrow \vec{a} = \frac{μ}{3 \sqrt{2}} (3 \hat{i} + 3 \hat{j} - \hat{i} + \hat{j} - 4 \hat{k})$
$= \frac{μ}{3 \sqrt{2}} (2 \hat{i} + 4 \hat{j} - 4 \hat{k})$
          Comparing with $\vec{a} = α \hat{i} + 2 \hat{j} + β \hat{k}$ , we get
     $\frac{4 μ}{3 \sqrt{2}} = 2 \Rightarrow μ = \frac{3 \sqrt{2}}{2}$
$\Rightarrow \vec{a} = \hat{i} + 2 \hat{j} - 2 \hat{k}$
Now $\vec{a} . \hat{k} + 2 = - 2 + 2 = 0$

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