AP EAMCET · Maths · Vector Algebra
If \(\overrightarrow{\mathrm{OA}}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{O B}=3 \hat{i}-\hat{k}\) and \(\overrightarrow{O C}=2 \hat{j}+3 \hat{k}\) are the position vectors of the points \(A, B\) and \(C\), then a unit vector perpendicular to the plane containing \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) is
- A \(\frac{8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{2 \sqrt{21}}\)
- B \(\frac{6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}}{7}\)
- C \(\frac{9 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}}{11}\)
- D \(\frac{8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{93}}\)
Answer & Solution
Correct Answer
(D) \(\frac{8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{93}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text {} \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\hat{i}+\hat{j}-2 \hat{k} \\ & \overrightarrow{A C}=-2 \hat{i}+3 \hat{j}+2 \hat{k} \end{aligned}\) \(\because \overrightarrow{A B}\) and \(\overrightarrow{A C}\) lies on the plane. Then…
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