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VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA3 Marks
Question
Show that the points A, B and C with position vectors, $\vec{a}=3\hat{i}-4\hat{j}-4\hat{k},\ \vec{b}=2\hat{i}-\hat{j}+\hat{k}\ $ $\text{and}\ \vec{c}=\hat{i}-3\hat{j}-5\hat{k},$ respectively form the vertices of a right angled triangle.
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Answer

Position vectors of points A, B, and C are respectively given as:
$\vec{a}=3\hat{i}-4\hat{j}-4\hat{k},\ \vec{b}=2\hat{i}-\hat{j}+\hat{k}\ \text{and}\ \vec{c}$ $=\hat{i}-3\hat{j}-5\hat{k}$
$\therefore\overrightarrow{\text{AB}}=\vec{b}-\vec{a}={(2-3)}\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}$ $=-\hat{i}+3\hat{j}+5\hat{k}$
$\overrightarrow{\text{BC}}=\vec{c}-\vec{b}={(1-2)}\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}$ $=-\hat{i}-2\hat{j}-6\hat{k}$
$\overrightarrow{\text{CA}}=\vec{a}-\vec{c}={(3-1)}\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}$ $=2\hat{i}-\hat{j}+\hat{k}$
$\therefore\bigg|\overrightarrow{\text{AB}}\bigg|^2=(-1)^2+3^2+5^2=1+9+25=35$
$\bigg|\overrightarrow{\text{BC}}\bigg|^2=(-1)^2+(-2)^2+(-6)^2$ $=1+4+36=41$
$\bigg|\overrightarrow{\text{CA}}\bigg|^2=2^2+(-1)^2+1^2=4+1+1=6$
$\therefore\bigg|\overrightarrow{\text{AB}}\bigg|^2+\bigg|\overrightarrow{\text{CA}}\bigg|^2=36+6=41=\bigg|\overrightarrow{\text{BC}}\bigg|^2$
Hence, ABC is a right-angled triangle.

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