Question

If the angle between the planes \( \lambda x-2y+3z+1=0 \) and \( 2x+3y-\lambda z+\lambda=0 \) is \( \cos^{-1}(\frac{12}{49}) \) and \( \lambda\in\mathbb{Z} \), then the sum of the perpendicular distances from the origin to these planes is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For angle-between-planes problems, always solve for the normal vectors first; once \( |\vec n_1| = |\vec n_2| \), the cosine expression simplifies drastically to a single-variable rational equation.

Answer 1

The Correct Option is A

Step 1: Write the normal vectors.
Plane 1 $\lambda x-2y+3z+1=0$ has normal $\vec{n_1}=(\lambda,-2,3)$. Plane 2 $2x+3y-\lambda z+\lambda=0$ has normal $\vec{n_2}=(2,3,-\lambda)$.
Step 2: Form the cosine of the angle.
The dot product is $\vec{n_1}\cdot\vec{n_2}=2\lambda-6-3\lambda=-\lambda-6$, and both lengths equal $\sqrt{\lambda^2+13}$. So \[ \cos\theta=\frac{|\lambda+6|}{\lambda^2+13} \]
Step 3: Use the given angle.
Setting this equal to $\frac{12}{49}$: \[ 49|\lambda+6|=12(\lambda^2+13) \]
Step 4: Find the integer $\lambda$.
Trying $\lambda=6$: left side $=49(12)=588$, right side $=12(36+13)=12(49)=588$. They match, so $\lambda=6$.
Step 5: Distance from the origin to each plane.
Distance equals $\dfrac{|\text{constant}|}{\sqrt{\text{sum of squares of }x,y,z\text{ coefficients}}}$. For plane 1 with constant $1$: \[ d_1=\frac{1}{\sqrt{36+4+9}}=\frac{1}{7} \] For plane 2 with constant $\lambda=6$: \[ d_2=\frac{6}{\sqrt{4+9+36}}=\frac{6}{7} \]
Step 6: Add the distances.
\[ d_1+d_2=\frac{1}{7}+\frac{6}{7}=1 \] \[ \boxed{1} \]