Q71 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
If $\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{k}+\hat{i}, \hat{i}-\hat{j}, \hat{j}-\hat{k}$ are the position vectors of the points $A, B, \dot{C}, D, E$ respectively, then the point of intersection of the line $A B$ and the plane passing through $C, D, E$ is
- A$\hat{i}+\hat{j}+\hat{k}$
- Correct answerB$\frac{1}{2} \hat{i}+\hat{j}+\frac{1}{2} \hat{k}$
- C$\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
- D$\frac{1}{2} \hat{i}-\hat{j}+\frac{1}{2} \hat{k}$
Solution
Equation of line $A B$
$\begin{aligned}
& \vec{r}=\hat{i}+\hat{j}+\lambda(-\hat{i}+\hat{k}) \\
& \Rightarrow \frac{x-1}{-1}=\frac{y-1}{0}=\frac{z-0}{1}=\lambda \\
& \Rightarrow x=-\lambda+1, y=\hat{1}, z=\lambda
\end{aligned}$
Equation of plane passing through $C, D$ and $E$ $\begin{aligned} & \left|\begin{array}{ccc} x-1 & y & z-1 \\ 0 & -1 & -1 \\ -1 & 1 & -2 \end{array}\right|=0 \\ & \Rightarrow(x-1)(2+1)-y(0-1)+(z-1)(0-1)=0 \\ & \Rightarrow 3(-\lambda+1-1)+1-(\lambda-1)=0 \\ & \Rightarrow-3 \lambda+1-\lambda+1=0 \Rightarrow-4 \lambda=-2 \Rightarrow \lambda=\frac{1}{2} \\ & \therefore \text { Point }=\frac{1}{2} \hat{i}+\hat{j}+\frac{1}{2} \hat{k} \end{aligned}$
Equation of plane passing through $C, D$ and $E$ $\begin{aligned} & \left|\begin{array}{ccc} x-1 & y & z-1 \\ 0 & -1 & -1 \\ -1 & 1 & -2 \end{array}\right|=0 \\ & \Rightarrow(x-1)(2+1)-y(0-1)+(z-1)(0-1)=0 \\ & \Rightarrow 3(-\lambda+1-1)+1-(\lambda-1)=0 \\ & \Rightarrow-3 \lambda+1-\lambda+1=0 \Rightarrow-4 \lambda=-2 \Rightarrow \lambda=\frac{1}{2} \\ & \therefore \text { Point }=\frac{1}{2} \hat{i}+\hat{j}+\frac{1}{2} \hat{k} \end{aligned}$
Hence, the correct answer is option: 'B'.