Question

The distance between a point \(P\) whose position vector is \[ 5\vec{i}+\vec{j}+3\vec{k} \] and the line \[ \vec{r}=(3\vec{i}+7\vec{j}+\vec{k})+t(\vec{j}+\vec{k}) \] is:

• A. \(3\)
• B. \(4\)
• C. \(5\)
• D. \(6\)

Hint:

For the distance between a point and a line in 3D, \[ d=\frac{|(\overrightarrow{AP}\times \vec{d})|}{|\vec{d}|}, \] where \(\vec{d}\) is the direction vector of the line and \(A\) is any point on the line.

Answer 1

The Correct Option is D

Step 1: Extract a point on the line and its direction vector.
From r⃗ = (3î + 7ĵ + k̂) + t(ĵ + k̂), a point on the line is A(3, 7, 1) and the direction vector is d⃗ = (0, 1, 1). The external point is P(5, 1, 3).

Step 2: Construct the vector from A to P.
AP⃗ = (5 - 3)î + (1 - 7)ĵ + (3 - 1)k̂ = 2î - 6ĵ + 2k̂.

Step 3: Apply the point-to-line distance formula.
Distance d = |AP⃗ × d⃗|/|d⃗|. Compute the cross product: AP⃗ × d⃗ = |î ĵ k̂; 2 -6 2; 0 1 1| = î(-6 - 2) - ĵ(2 - 0) + k̂(2 - 0) = -8î - 2ĵ + 2k̂.

Step 4: Calculate the magnitudes.
|AP⃗ × d⃗| = √(64 + 4 + 4) = √72 = 6√2. |d⃗| = √(0² + 1² + 1²) = √2.

Step 5: Compute the distance.
d = (6√2)/√2 = 6.

Step 6: Final conclusion.
The perpendicular distance is 6 units.