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If the orthocentre of the triangle whose vertices are $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$, then
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Verified Answer
The correct answer is:
$x=y=z$
The vertices of triangle are
$2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$\mathbf{A B}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\mathbf{B C}=-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\mathbf{A C}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
$|\mathbf{A B}|=\sqrt{9+1+4}=\sqrt{14}$
$|\mathbf{B C}|=\sqrt{4+9+1}=\sqrt{14}$
$|\mathbf{A C}|=\sqrt{1+4+9}=\sqrt{14}$
$\because \triangle A B C$ is an equilateral triangle
$\therefore$ Orthocentre of triangle is coincide with centroid
$\therefore$ Orthocentre of $\triangle A B C$ is
$\left(\frac{2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}+5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{3}\right)$
$=\frac{15 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}}{3}=5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
$x=y=z=5$
$2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$\mathbf{A B}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\mathbf{B C}=-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\mathbf{A C}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
$|\mathbf{A B}|=\sqrt{9+1+4}=\sqrt{14}$
$|\mathbf{B C}|=\sqrt{4+9+1}=\sqrt{14}$
$|\mathbf{A C}|=\sqrt{1+4+9}=\sqrt{14}$
$\because \triangle A B C$ is an equilateral triangle
$\therefore$ Orthocentre of triangle is coincide with centroid
$\therefore$ Orthocentre of $\triangle A B C$ is
$\left(\frac{2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}+5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{3}\right)$
$=\frac{15 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}}{3}=5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
$x=y=z=5$
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