The position vectors of the points A, B, C are 2 i + j - k , 3 i -2 j + k and i +4 j -3 k respectively. These points,

The position vectors of the points A, B, C are 2 i + j - k , 3 i …

MCQ

The position vectors of the points A, B, C are $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ respectively. These points,

✓
Form an isosceles triangle.
B
Form a right triangle.
C
Are collinear.
D
Form a scalene triangle.

✓

Answer

Correct option: A.

Form an isosceles triangle.

Given: Position vectors of A, B, C are $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$. Then,

$\overrightarrow{\text{AB}}=\big(3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)-\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$

$=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$

$\overrightarrow{\text{BC}}=\big(\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)-\big(3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$

$=-2\hat{\text{i}}+6\hat{\text{j}}-4\hat{\text{k}}$

$\overrightarrow{\text{CA}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)-\big(\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)$

$=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$

Now, $\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{1^2+(-3)^2+2^2}$

$=\sqrt{1+9+4}$

$=\sqrt{14}$

$\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{1^2+(-3)^2+2^2}$

$=\sqrt{1+9+4}$

$=\sqrt{14}$

$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(-2)^2+6^2+(-4)^2}$

$=\sqrt{4+36+16}$

$=\sqrt{56}$

$\therefore\Big|\overrightarrow{\text{AB}}\Big|=\Big|\overrightarrow{\text{CA}}\Big|$

Hence, the triangle is isosceles as two of its sides are equal.