Question

Let \( \lambda_1 < \lambda_2 \) be two values of \( \lambda \) such that the angle between the planes:
\( P_1 : \vec{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7 \) 
\( P_2 : \vec{r} \cdot (\lambda\hat{i} + \hat{j} - 3\hat{k}) = 9 \)
is \( \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right) \). Then, the square of the length of the perpendicular from the point \( (38\lambda, 10\lambda, 2) \) to the plane \( P_1 \) is:

Answer 1

Correct Answer: 315

The two planes are given as:

\[ P_1 : \vec{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7 \] \[ P_2 : \vec{r} \cdot (\lambda\hat{i} + \hat{j} - 3\hat{k}) = 9 \]

Angle Between the Planes

The angle \( \theta \) between the planes is given by:

\[ \sin \theta = \frac{|\vec{n}_1 \times \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|} \]

where \( \vec{n}_1 = \langle 3, -5, 1 \rangle \) and \( \vec{n}_2 = \langle \lambda, 1, -3 \rangle \).

The magnitude of the cross product is:

\[ \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -5 & 1 \\ \lambda & 1 & -3 \end{vmatrix} = -i + 7j + 11k \]

From this:

\[ \sin \theta = \frac{|2\sqrt{6}|}{5} \]

Evaluate \( \cos \theta \)

We are given \( \cos \theta = \frac{1}{5} \). Square both sides and simplify:

\[ \cos \theta = \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|} \] \[ \cos \theta = \frac{3\lambda - 8}{\sqrt{35} \cdot \sqrt{\lambda^2 + 10}} \]

Substitute \( \cos \theta = \frac{1}{5} \):

\[ \frac{(3\lambda - 8)^2}{35(\lambda^2 + 10)} = \frac{1}{25} \]

Multiply through and simplify:

\[ 19\lambda^2 - 120\lambda + 125 = 0 \]

Solve the Quadratic Equation

Factorize:

\[ 19\lambda^2 - 95\lambda - 25\lambda + 125 = 0 \] \[ \lambda = 5, \, \lambda = \frac{25}{19} \]

Perpendicular Distance of a Point from Plane

The point \( \vec{r} = (38\lambda, 10\lambda, 2) \) is substituted into plane \( P_1 \). For \( \lambda = 5 \), the coordinates become \( (50, 50, 2) \).

The perpendicular distance from \( P_1 \) is:

\[ \frac{|3 \cdot 50 - 5 \cdot 50 + 2 - 7|}{\sqrt{35}} = \frac{105}{\sqrt{35}} \]

Square the result:

\[ \left(\frac{105}{\sqrt{35}}\right)^2 = 315 \]

Conclusion

The final result is \( 315 \).