Question

Let A be a point on the line \(\vec{r} = (1-3\mu)\hat{i} + (\mu-1)\hat{j} + (2+5\mu)\hat{k}\) and B(3, 2, 6) be a point in the space. Then the value of \(\mu\) for which the vector \(\overrightarrow{AB}\) is parallel to the plane \(x - 4y + 3z = 1\) is -

• A. \(\frac{1}{8}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{4}\)
• D. \(-\frac{1}{4}\)

Hint:

A vector parallel to a plane is perpendicular to the plane's normal \((1,-4,3)\); use the dot product.

Answer 1

The Correct Option is C

A \(=(1-3\mu,\ \mu-1,\ 2+5\mu)\) and B \(=(3,2,6)\).
\(\overrightarrow{AB} = (3-(1-3\mu),\ 2-(\mu-1),\ 6-(2+5\mu)) = (2+3\mu,\ 3-\mu,\ 4-5\mu)\).
The normal of the plane is \(\vec n = (1,-4,3)\). For \(\overrightarrow{AB}\) parallel to the plane, \(\overrightarrow{AB}\cdot\vec n = 0\).
\((2+3\mu) - 4(3-\mu) + 3(4-5\mu) = 2+3\mu-12+4\mu+12-15\mu = 2-8\mu = 0\).
So \(\mu = \frac14\).
Answer: (3) \(\frac{1}{4}\)