Question

For a positive real number p, if the perpendicular distance from a point \( -\vec{i} + p\vec{j} - 3\vec{k} \) to the plane \( \vec{r} \cdot (2\vec{i} - 3\vec{j} + 6\vec{k}) = 7 \) is 6 units, then p =

• A. \( \frac{4}{5} \)
• B. \( \frac{5}{6} \)
• C. \( 6 \)
• D. \( 5 \) Correct Answer

Hint:

The perpendicular distance from a point with position vector \( \vec{a} \) to the plane \( \vec{r} \cdot \vec{n} = d_0 \) is given by \( D = \frac{|\vec{a} \cdot \vec{n} - d_0|}{|\vec{n}|} \). Alternatively, for a point \( (x_1, y_1, z_1) \) and a plane \( Ax+By+Cz+D_{cartesian}=0 \), the distance is \( \frac{|Ax_1+By_1+Cz_1+D_{cartesian}|}{\sqrt{A^2+B^2+C^2}} \). Note that \( d_0 \) in vector form is usually on the RHS, while \(D_{cartesian}\) for the Cartesian formula is on the LHS. If the plane is \( Ax+By+Cz=d_0 \), the formula using Cartesian coordinates is \( \frac{|Ax_1+By_1+Cz_1-d_0|}{\sqrt{A^2+B^2+C^2}} \).

Answer 1

The Correct Option is D

Step 1: Identify the point and the plane equation.
The point is \( \vec{a} = -\vec{i} + p\vec{j} - 3\vec{k} = (-1, p, -3) \).
The plane equation is \( \vec{r} \cdot \vec{n} = d_0 \), where \( \vec{n} = 2\vec{i} - 3\vec{j} + 6\vec{k} \) and \( d_0 = 7 \).
The Cartesian form of the plane is \( 2x - 3y + 6z = 7 \), or \( 2x - 3y + 6z - 7 = 0 \).

Step 2: Use the formula for the perpendicular distance from a point to a plane.
The perpendicular distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax+By+Cz+D'=0 \) is given by: \[ D = \frac{|Ax_0+By_0+Cz_0+D'|}{\sqrt{A^2+B^2+C^2}} \] In vector form, the distance from point \( \vec{a} \) to plane \( \vec{r} \cdot \vec{n} = d_0 \) is: \[ D = \frac{|\vec{a} \cdot \vec{n} - d_0|}{|\vec{n}|} \]
Step 3: Calculate \( \vec{a} \cdot \vec{n} \) and \( |\vec{n}| \).
\( \vec{a} \cdot \vec{n} = (-1)(2) + (p)(-3) + (-3)(6) = -2 - 3p - 18 = -20 - 3p \).
\( |\vec{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7 \).

Step 4: Substitute into the distance formula and set it equal to the given distance.
The given distance is \( D=6 \) units.
\[ 6 = \frac{|(-20-3p) - 7|}{7} = \frac{|-27-3p|}{7} \] \[ 6 = \frac{|-(27+3p)|}{7} = \frac{|27+3p|}{7} \]
Step 5: Solve for p.
\[ 7 \times 6 = |27+3p| \] \[ 42 = |27+3p| \] This gives two possibilities: Case 1: \( 27+3p = 42 \) \[ 3p = 42 - 27 = 15 \] \[ p = \frac{15}{3} = 5 \] Case 2: \( 27+3p = -42 \) \[ 3p = -42 - 27 = -69 \] \[ p = \frac{-69}{3} = -23 \]
Step 6: Apply the condition that p is a positive real number.
Since p must be positive, we choose \( p=5 \).
This matches option (4).