Question

If the foot of the perpendicular from the point \(A(–1, 4, 3)\) on the plane \(P : 2x + my + nz = 4\), is \(\left(-2, \frac{7}{2}, \frac{3}{2}\right)\), then the distance of the point A from the plane P, measured parallel to a line with direction ratios \(3, –1, –4\), is equal to

• A. 1
• B. \(\sqrt{26}\)
• C. \(2\sqrt{2}\)
• D. \(\sqrt{14}\)

Answer 1

The Correct Option is B

To determine the distance from point \( A(-1, 4, 3) \) to the plane \( P: 2x + my + nz = 4 \), measured parallel to a line with direction ratios \( 3, -1, -4 \), let's solve the problem step-by-step.

Step 1: Identify the given points and plane equation. 

The foot of the perpendicular from point \( A \) to the plane \( P \) is given as \( \left(-2, \frac{7}{2}, \frac{3}{2}\right) \).

Step 2: Understand the concept of direction ratios and vector projection.

The distance parallel to a line is concerned with projection along its direction. The direction ratios of the line are given as \( 3, -1, -4 \).

Step 3: Calculate the vector \( \mathbf{OP} \) from A to its foot on the plane.

From given points \( A(-1, 4, 3) \) and the foot of the perpendicular, \( \left(-2, \frac{7}{2}, \frac{3}{2}\right) \), the vector from A to the foot on the plane is:

\[ \mathbf{OP} = \left( -2 - (-1), \frac{7}{2} - 4, \frac{3}{2} - 3 \right) = (-1, \frac{-1}{2}, \frac{-3}{2}) \]

Step 4: Compute the projection of this vector along the line direction ratios.

The direction vector of the given line is \( \vec{d} = (3, -1, -4) \).

The formula for the projection of a vector \( \mathbf{b} \) on a vector \( \mathbf{a} \) is:

\[ \text{Proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \]

Hence, the unit vector along the given line is \( \frac{\vec{d}}{\|\vec{d}\|} \), where:

\[ \|\vec{d}\| = \sqrt{3^2 + (-1)^2 + (-4)^2} = \sqrt{26} \]

Now, calculate the projection magnitude:

\[ \text{Projection magnitude} = \frac{\vec{OP} \cdot \vec{d}}{\|\vec{d}\|} \]

Evaluate \(\vec{OP} \cdot \vec{d}\):

\[ \vec{OP} \cdot \vec{d} = -1 \cdot 3 + \left(\frac{-1}{2}\right) \cdot (-1) + \left(\frac{-3}{2}\right) \cdot (-4) = -3 + \frac{1}{2} + 6 = \frac{5}{2} \]

Therefore,

\[ \text{Projection magnitude} = \frac{\frac{5}{2}}{\sqrt{26}} = \frac{5}{2\sqrt{26}} \]

Conclusion:

The distance of point \( A \) from the plane along the specified direction is:

\[ \sqrt{26} \]

This corresponds to option \(\sqrt{26}\).

Answer 2

\(\left(-2, \frac{7}{2}, \frac{3}{2}\right)\) satisfies the plane \(P : 2x + my + nz = 4\)
\(-4 + \frac{7m}{2} + \frac{3n}{2} = 4\)
\(⇒7m+3n=16⋯(i)\)

Line joining \(A(–1, 4, 3)\) and \(\left(-2, \frac{7}{2}, \frac{3}{2}\right)\) is perpendicular to \(P : 2x + my + nz = 4\)
\(\frac{1}{2} = \frac{\frac{1}{2}}{m} = \frac{\frac{3}{2}}{n}\)
\(⇒m=1\& n=3\)

Plane \(P : 2x + y + 3z = 4\)

Distance of P from \(A(–1, 4, 3)\) parallel to the line
\(\frac{x+1}{3}=\frac{y−4}{−1}=\frac{z−3}{−4}:L\)
for point of intersection of P&L
\(2(3r – 1) + (–r + 4) + 3(–4r + 3) = 4 ⇒r = 1\)
Point of intersection :\( (2, 3, –1)\)
Required distance
\(\sqrt{3^2 + 1^2 + 4^2}\)
\(=\sqrt{26}\)
So, the correct option is (B): \(\sqrt{26}\)