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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Vector Algebra

BITSAT 2023 · Paper 2 · Mathematics · Vector Algebra

A vector perpendicular to the plane containing the vectors i -2 j -...

A vector perpendicular to the plane containing the vectors i^−2j^−k^\hat{i}-2 \hat{j}-\hat{k} and 3i^−2j^−k^3 \hat{i}-2 \hat{j}-\hat{k} is inclined to the vector i^+j^+k^\hat{i}+\hat{j}+\hat{k} at an angle

Options

  1. Atan⁡−114\tan ^{-1} \sqrt{14}
  2. Bsec⁡−114\sec ^{-1} \sqrt{14}
  3. Ctan⁡−115\tan ^{-1} \sqrt{15}
  4. D
    None of these

Correct answer

A. tan⁡−114\tan ^{-1} \sqrt{14}

Step-by-step solution

A vector perpendicular to the plane is(i^−2j^−k^)×(3i^−2j^−k^)=∣i^j^k^1−2−13−2−1∣=−2j^+4k^⇒ unit vector a^=−2j^+4k^4+16=−2j^+4k^25\begin{aligned} \\ & (\hat{i}-2 \hat{j}-\hat{k}) \times(3 \hat{i}-2 \hat{j}-\hat{k}) \\ \\ & =\left|\begin{array}{ccc} \\ \hat{i} & \hat{j} & \hat{k} \\ \\ 1 & -2 & -1 \\ \\ 3 & -2 & -1 \\ \end{array}\right|=-2 \hat{j}+4 \hat{k} \\ \\ & \Rightarrow \text{ unit vector } \hat{a}=\frac{-2 \hat{j}+4 \hat{k}}{\sqrt{4+16}}=\frac{-2 \hat{j}+4 \hat{k}}{2 \sqrt{5}} \\ \end{aligned}Angle between the unit vector and r⃗=i^+j^+k^=cos⁡−1r⃗⋅a^∣r⃗∣⋅∣a^∣=cos⁡−1115=tan⁡−114\vec{r}=\hat{i}+\hat{j}+\hat{k} =\cos ^{-1} \frac{\vec{r} \cdot \hat{a}}{|\vec{r}| \cdot|\hat{a}|}=\cos ^{-1} \frac{1}{\sqrt{15}}=\tan ^{-1} \sqrt{14}
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