Question

The value of $a$, for which the points $A,B$ and $C$ with position vectors $2i-j+k$ and $i-3j-5k$ and $ai-3j+k$ respectively are the vertices of a right-angled triangle with $C=\frac{\mathrm{\pi }}{2}$ are?

• A. $-2$ and $-1$
• B. $-2$ and $1$
• C. $2$ and $-1$
• D. $2$ and $1$

Answer 1

The correct option is D

$2$ and $1$

Explanation for the correct option:

Find the value of $a$:

In the question, it is given that the position vector of points $A,B$ and $C$ are $2i-j+k$ and $i-3j-5k$ and $ai-3j+k$ respectively and the vertices of a right-angled triangle with $C=\frac{\mathrm{\pi }}{2}$.

Since, it is given that $C=\frac{\mathrm{\pi }}{2}$.

We know that the dot product of perpendicular vectors is zero.

Therefore, $\overrightarrow{AC}·\overrightarrow{BC}=0...\left(1\right)$.

Now, find $\overrightarrow{AC}$.

$$\begin{gathered} \overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A} \\ \Rightarrow \overrightarrow{AC}=(ai-3j+k)-(2i-j+k) \\ \Rightarrow \overrightarrow{AC}=(a-2)i-2j \end{gathered}$$

Find $\overrightarrow{BC}$.

$$\begin{gathered} \overrightarrow{BC}=\overrightarrow{C}-\overrightarrow{B} \\ \Rightarrow \overrightarrow{BC}=(ai-3j+k)-(i-3j-5k) \\ \Rightarrow \overrightarrow{BC}=(a-1)i+6k \end{gathered}$$

From equation $\left(1\right)$.

$$\begin{gathered} \left[\right(a-2)i-2j]·\left[\right(a-1)i+6k]=0 \\ \Rightarrow (a-2)(a-1)+0+0+0=0 \\ \Rightarrow a=1,2 \end{gathered}$$

Therefore, The values of $a$, for which the points $A,B$ and $C$ with position vectors $2i-j+k$ and $i-3j-5k$ and $ai-3j+k$ respectively are the vertices of a right-angled triangle with $C=\frac{\mathrm{\pi }}{2}$ are $1,2$.

Hence, option (D), $1,2$ is the correct answer.