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If $\mathbf{a}$ and $\mathbf{b}$ are respectively the internal and external bisectors of the angles between the vectors $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ and $|\mathbf{a}|=\frac{2}{3} \sqrt{6},|\mathbf{b}|=\frac{2}{3} \sqrt{3}$, then one of the values of $\mathbf{a}-\mathbf{b}$ is
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Verified Answer
The correct answer is:
$\frac{2}{3}(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$
We have,
$-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \text { and } 3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}
$$
$\mathbf{a}$ and $\mathbf{b}$ are the angle bisector of vectors
$$
\begin{aligned}
& \therefore|\mathbf{a}|=\lambda\left|\frac{3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}}{5}+\frac{-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{3}\right| \\
& =\lambda\left|\frac{4 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}}{15}\right| \\
& |\mathbf{a}|=\frac{2}{3} \sqrt{6}=\frac{\lambda}{15} \sqrt{16+484+100} \\
& \Rightarrow \quad \lambda=1 \\
& \text { Similarly }|\mathbf{b}|=\mu\left[\frac{3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}}{5}-\frac{-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{3}\right] \\
& =\mu\left(\frac{14 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}}{15}\right) \\
& |\mathbf{b}|=\frac{2}{3} \sqrt{3}=\frac{\mu}{15} \sqrt{196+4+100} \\
& \Rightarrow \quad \mu=1 \\
& \therefore \quad \mathbf{a}-\mathbf{b}=\frac{4 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}}{15}-\frac{14 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}}{15} \\
& =\frac{2}{3}(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\
&
\end{aligned}$
$-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \text { and } 3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}
$$
$\mathbf{a}$ and $\mathbf{b}$ are the angle bisector of vectors
$$
\begin{aligned}
& \therefore|\mathbf{a}|=\lambda\left|\frac{3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}}{5}+\frac{-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{3}\right| \\
& =\lambda\left|\frac{4 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}}{15}\right| \\
& |\mathbf{a}|=\frac{2}{3} \sqrt{6}=\frac{\lambda}{15} \sqrt{16+484+100} \\
& \Rightarrow \quad \lambda=1 \\
& \text { Similarly }|\mathbf{b}|=\mu\left[\frac{3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}}{5}-\frac{-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{3}\right] \\
& =\mu\left(\frac{14 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}}{15}\right) \\
& |\mathbf{b}|=\frac{2}{3} \sqrt{3}=\frac{\mu}{15} \sqrt{196+4+100} \\
& \Rightarrow \quad \mu=1 \\
& \therefore \quad \mathbf{a}-\mathbf{b}=\frac{4 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}}{15}-\frac{14 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}}{15} \\
& =\frac{2}{3}(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\
&
\end{aligned}$
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