Question

Let the line / : x = \(\frac{1-y}{2}=\frac{z-3}{\lambda}, \lambda \in R\) meet the plane P : x+2y +3z = 4 at the point \((\alpha, \beta, \lambda)\). If the angle between the line I and the plane P is \(cos^{-1}\bigg(\sqrt{\frac{5}{14}}\bigg)\) then \(\alpha+2\beta+6\lambda\) is equal to______.

Answer 1

Correct Answer: 11

The given line \( \ell \) is:

\( \frac{y - 1}{2} = \frac{z - 3}{\lambda}, \, \lambda \in \mathbb{R}. \)

Step 1: Direction ratios of the line

The direction ratios (Dr’s) of the line \( \ell \) are \( (1, 2, \lambda) \).

Step 2: Direction ratios of the normal vector of the plane

The equation of the plane is:

\( x + 2y + 3z = 4. \)

The direction ratios of the normal vector of the plane are \( (1, 2, 3) \).

Step 3: Angle between the line and the plane

The angle between the line \( \ell \) and the plane \( P \) is given by:

\( \sin \theta = \frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}}. \)

It is given that:

\( \cos \theta = \frac{5}{14}, \, \text{so} \, \sin \theta = \frac{3\sqrt{14}}{14}. \)

Equating:

\( \frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}} = \frac{3\sqrt{14}}{14}. \)

Simplify:

\( |1 + 4 + 3\lambda| = 3 \implies 1 + 4 + 3\lambda = 3. \)

Solve for \( \lambda \):

\( 3\lambda = 2 \implies \lambda = \frac{2}{3}. \)

Step 4: Variable point on the line

The parametric equation of the line \( \ell \) is:

\( (t, 2t + 1, \frac{2}{3}t + 3). \)

Step 5: Find the intersection point with the plane

The point lies on the plane:

\( t + 2(2t + 1) + 3\left(\frac{2}{3}t + 3\right) = 4. \)

Simplify:

\( t + 4t + 2 + 2t + 9 = 4. \)

\( 7t + 11 = 4 \implies t = -1. \)

Substitute \( t = -1 \) into the parametric equation of the line:

\( (-1, -1, \frac{7}{3}) = (\alpha, \beta, \gamma). \)

Step 6: Verify the condition

The point satisfies:

\( \alpha + 2\beta + 6\gamma = -1 + 2(-1) + 6\left(\frac{7}{3}\right). \)

Simplify:

\( -1 - 2 + 14 = 11. \)

Final Answer:

The point is \( (-1, -1, \frac{7}{3}) \), and the condition \( \alpha + 2\beta + 6\gamma = 11 \) is satisfied.