Question

Let P (3,2,6) be a point in space and Q be a point on the line \[{\vec r } = {\text{ }}\left( {\hat i - \hat j + 2\hat k} \right){\text{ }} + {\text{ }}\mu \left( { - 3\hat i + \hat j + 5\hat k} \right)\] . Then the value of $\mu $ for which the vector PQ is parallel to the plane x-4y+3z = 1 is:
$
  {\text{A}}{\text{. }}\dfrac{1}{4} \\
  {\text{B}}{\text{. - }}\dfrac{1}{4} \\
  {\text{C}}{\text{. }}\dfrac{1}{8} \\
  {\text{D}}{\text{. - }}\dfrac{1}{4} \\
$

Answer 1

Hint: Assume the coordinates of the point Q on the line. Now determine the direction ratios of vector PQ and then take dot product of these direction ratios with those of the plane and equate it to 0 to get the value of $\mu $.

Complete step-by-step answer:
The coordinates of point P are given as (3,2,6).
Any point Q lies on the line \[{\vec r } = {\text{ }}\left( {\hat i - \hat j + 2\hat k} \right){\text{ }} + {\text{ }}\mu \left( { - 3\hat i + \hat j + 5\hat k} \right)\] where $\mu $ is a constant and this equation of line can also be written as:
\[{\vec r } = {\text{ }}\left( {(1 - 3\mu )\hat i + ( - 1 + \mu )\hat j + (2 + 5\mu )\hat k} \right)\]
So, we can take the coordinates of Q as\[\left( {(1 - 3\mu ),( - 1 + \mu ),(2 + 5\mu )} \right)\]
Therefore, the direction ratios of PQ are:
$
  \left( {(1 - 3\mu - 3),( - 1 + \mu - 2),(2 + 5\mu - 6)} \right) \\
  \left( {( - 3\mu - 2),(\mu - 3),(5\mu - 4)} \right) \\
$
The equation of the plane given is x-4y+3z = 1.
The direction ratios of the plane are (1,-4,3).
Now it is mentioned in the question that the vector PQ is parallel to the plane x-4y+3z = 1, therefore the vector PQ must be perpendicular to the normal of the plane.
Therefore, the dot product of direction ratios of the vector PQ and the normal of the plane must be equal to 0.
Therefore,
$
  \left( {1( - 3\mu - 2) - 4(\mu - 3) + 3(5\mu - 4)} \right) = 0 \\
   - 2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0 \\
  8\mu - 2 = 0 \\
  \mu = \dfrac{1}{4} \\
$
Hence the value of $\mu $ is obtained to be $\dfrac{1}{4}$ .
The correct option is A.

Note: The dot product of the direction ratios of the line vector PQ and the plane were 0, since the direction ratios of the plane basically are the D.Rs of its normal. And any line parallel to the plane will be perpendicular with the normal of the plane. So the dot product is 0.