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MHT CET · Maths · Three Dimensional Geometry

Let \(\bar{A}\) be a vector parallel to line of intersection of planes \(P_1\) and \(P_2\) through origin. \(P_1\) is parallel to the vectors \(2 \hat{j}+3 \hat{k}\) and \(4 \hat{j}-3 \hat{k}\) and \(P_2\) is parallel to \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\), then the angle between \(\bar{A}\) and \(2 \hat{i}+\hat{j}-2 \hat{k}\) is

  1. A \(\frac{\pi}{3}\)
  2. B \(\frac{\pi}{2}\)
  3. C \(\frac{\pi}{6}\)
  4. D \(\frac{3 \pi}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{3 \pi}{4}\)

Step-by-step Solution

Detailed explanation

Vector equation of the plane passing through the point \(A(\bar{a})\) and parallel to non-zero vectors \(\bar{b}\) and \(\overline{\mathrm{c}}\) is \(\overline{\mathrm{r}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})\)
Plane \(P_1\) is passing through the origin and parallel to vectors \(\overline{b_1}=2 \hat{j}+3 \hat{k}\) and \(\overline{c_1}=4 \hat{j}-3 \hat{k}\)
\(
\therefore \overline{\mathrm{b}_1} \times \overline{\mathrm{c}_1}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
0 & 2 & 3 \\
0 & 4 & -3
\end{array}\right|=-18 \hat{\mathrm{i}}
\)
\(\therefore\) Equation of \(\mathrm{P}_1\) is: \(\mathrm{r} \cdot(-18 \mathrm{i})=0\)
Plane \(P_2\) is passing through the origin and parallel to vectors \(\overline{b_2}=\hat{j}-\hat{k}\) and \(\overline{c_2}=3 \hat{i}+3 \hat{j}\)
\(
\therefore \overline{b_2} \times \overline{c_2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & -1 \\
3 & 3 & 0
\end{array}\right|=3 \hat{i}-3 \hat{j}-3 \hat{k}
\)
\(\therefore\) Equation of \(\mathrm{P}_2\) is \(: \mathrm{r}_2 \cdot(3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=0\)
Note that \(\overline{\mathrm{A}}\) is parallel to the cross product of \(-18 \hat{i}\) and \(3 \hat{i}-3 \hat{j}-3 \hat{k}\)
\(
\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-18 & 0 & 0 \\
3 & -3 & -3
\end{array}\right|=-54 \hat{\mathrm{j}}+54 \hat{\mathrm{k}}
\)
Let \(\theta\) be the required angle.
\(\therefore \theta=\) Angle between \(54(-\hat{\mathrm{j}}+\hat{\mathrm{k}})\) and \(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
\(\therefore \cos \theta =\frac{54 \times(-1-2)}{54 \sqrt{0+1+1} \sqrt{4+1+4}} \)
\( = \pm \frac{3}{3 \sqrt{2}} \)
\( = \pm \frac{1}{\sqrt{2}} \)
\( \therefore \theta=\frac{\pi}{4} , \frac{3 \pi}{4}\)