First slide

Algebra of vectors

Question

Let ABC be a triangle, the position vectors of whose vertices are $7 \hat{j} + 10 \hat{k}, - \hat{i} + 6 \hat{j} + 6 \hat{k}$ and $- 4 \hat{i} + 9 \hat{j} + 6 \hat{k}$ then $∆$ ABC is

Easy

A

isosceles
B

equilateral
C

right angled
D

none of these

Solution

we have $\vec{AB} = - \hat{i} - \hat{j} - 4 \hat{k}, \vec{BC} = - 3 \hat{i} + 3 \hat{j}$ and $\vec{CA} = 4 \hat{i} - 2 \hat{j} + 4 \hat{k}$
Therefore, $| \vec{A B} | = | \vec{B C} | = 3 \sqrt{2}$
clearly , $| \vec{AB} |^{2} + | \vec{BC} |^{2} = | \vec{AC} |^{2}$
Hence, the triangle is right-angled isosceles triangle.

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