Q74 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics
If $2 \vec{i}-\hat{j}+3 \hat{k},-12 \hat{i}-\hat{j}-3 \hat{k},-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then $\lambda=$
- A$9$
- B$-2$
- C$8$
- Correct answerD$6$
Solution
$\mathrm{A}: 2 \hat{i}-\hat{j}+3 \hat{k} ; \mathrm{B}:-12 \hat{i}-\hat{j}-3 \hat{k}$
$\begin{aligned}<br />& \mathrm{C}:-\hat{i}+2 \hat{j}-4 \hat{k} ; \mathrm{D}: \lambda \hat{i}+2 \hat{j}-\hat{k} \\<br />& \overrightarrow{\mathrm{AB}}=-14 \hat{i}-6 \hat{k} ; \overrightarrow{\mathrm{BC}}=11 \hat{i}+3 \hat{j}-\hat{k} \\<br />& \overrightarrow{\mathrm{AC}}=-3 \hat{i}+3 \hat{j}-7 \hat{k} ; \overrightarrow{\mathrm{AD}}=(\lambda-2) \hat{i}+3 \hat{j}-4 \hat{k}<br />\end{aligned}$
for coplanar vectors,
$\begin{aligned}<br />& \overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AD}})=0 \\<br />& \left|\begin{array}{ccc}<br />-14 & 0 & -6 \\<br />-3 & 3 & -7 \\<br />\lambda-2 & 3 & -4<br />\end{array}\right|=0 \\<br />& \Rightarrow 9(-14+2(3+\lambda-2))=0 \therefore \lambda=6<br />\end{aligned}$
$\begin{aligned}<br />& \mathrm{C}:-\hat{i}+2 \hat{j}-4 \hat{k} ; \mathrm{D}: \lambda \hat{i}+2 \hat{j}-\hat{k} \\<br />& \overrightarrow{\mathrm{AB}}=-14 \hat{i}-6 \hat{k} ; \overrightarrow{\mathrm{BC}}=11 \hat{i}+3 \hat{j}-\hat{k} \\<br />& \overrightarrow{\mathrm{AC}}=-3 \hat{i}+3 \hat{j}-7 \hat{k} ; \overrightarrow{\mathrm{AD}}=(\lambda-2) \hat{i}+3 \hat{j}-4 \hat{k}<br />\end{aligned}$
for coplanar vectors,
$\begin{aligned}<br />& \overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AD}})=0 \\<br />& \left|\begin{array}{ccc}<br />-14 & 0 & -6 \\<br />-3 & 3 & -7 \\<br />\lambda-2 & 3 & -4<br />\end{array}\right|=0 \\<br />& \Rightarrow 9(-14+2(3+\lambda-2))=0 \therefore \lambda=6<br />\end{aligned}$
Hence, the correct answer is option: 'D'.