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A vector \vec{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}\; (\alpha ,\beta \epsilon 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:
Option: 1 \vec{a}\cdot \hat{i}+3=0
Option: 2 \vec{a}\cdot \hat{k}+4=0
Option: 3 \vec{a}\cdot \hat{i}+1=0
Option: 4  \vec{a}\cdot \hat{k}+2=0
 

Answers (1)

best_answer

 

 

Equation of The Plane Bisecting the Angle Between Two Planes -

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angle bisector can be \vec{a}=\vec{a}=\lambda(\hat{b}+\hat{c}) \text { or } \vec{a}=\mu(\hat{b}-\hat{c})

\\\overrightarrow{\mathrm{a}}=\lambda\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}+\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{3 \sqrt{2}}\right)=\frac{\lambda}{3 \sqrt{2}}[3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}]\\=\frac{\lambda}{3 \sqrt{2}}[4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}]

Compare it with \overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}

\\ {\frac{2 \lambda}{3 \sqrt{2}}=2 \Rightarrow \lambda=3 \sqrt{2}} \\ {\overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}} \\ {\text { Not in option so now consider }} \overrightarrow{\mathrm{a}}=\mu\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}-\frac{\hat{i}-\hat{j}+4 \hat{k}}{3 \sqrt{2}}\right)

\begin{aligned} \overrightarrow{\mathrm{a}}=& \frac{\mu}{3 \sqrt{2}}(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{i}}+\hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \\=& \frac{\mu}{3 \sqrt{2}}(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \end{aligned}

Compare it with \overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}

\\ {\frac{4 \mu}{3 \sqrt{2}}=2 \Rightarrow \mu=\frac{3 \sqrt{2}}{2}} \\ {\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}} \\ {\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{k}}+2=0} \\ {-2+2=0}

Correct Option (4)

Posted by

Ritika Jonwal

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