Question

In a triangle $ABC,$ right angled at the vertex $A,$ if the position vectors of $A,B$ and $C$ are respectively $3\hat{i}+\hat{j}–\hat{k},–\hat{i}+3\hat{j}+p\hat{k}$ and $5\hat{i}+q\hat{j}–4\hat{k}$ then the point $(p,q)$ lies on a line

• A. making an acute angle with the positive direction of x-axis
• B. parallel to y-axis
• C. parallel to x-axis
• D. making an obtuse with the positive direction of x-axis

Answer 1

The correct option is A

making an acute angle with the positive direction of x-axis

Given: $A\equiv 3\hat{i}+\hat{j}–\hat{k},B\equiv –\hat{i}+3\hat{j}+p\hat{k}$ and $C\equiv 5\hat{i}+q\hat{j}–4\hat{k}$

$\Rightarrow AB=–4\hat{i}+2\hat{j}+(p+1)\hat{k}$ and $AC=2\hat{i}+(q–1)\hat{j}–3\hat{k}$

As $\angle CAB=\frac{\pi }{2}$ So, $AB.AC=0$

$\Rightarrow -4\times 2+2\times (q-1)+(p+1)\times (-3)=0$

$\Rightarrow -3p+2q-13=0$

Thus, $(p,q)$ lies on $3x-2y+13=0$ which makes an acute angle with the x-axis.