JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 3)

Then the value of $$\mu $$ for which the vector $${\overrightarrow {PQ} }$$ is parallel to the plane $$x - 4y + 3z = 1$$ is :

$${1 \over 4}$$

$$-{1 \over 4}$$

$${1 \over 8}$$

$$-{1 \over 8}$$

We see that any point on the line is given as

$$Q \equiv \{ (1 - 3\mu ),(\mu - 1),(5\mu + 2)\} $$

$$\overrightarrow {PQ} \equiv \{ - 3\mu - 2,\mu - 3,5\mu - 4\} $$

Now, we have

$$1( - 3\mu - 2) - 4(\mu - 3) + 3(5\mu - 4) = 0$$

$$ \Rightarrow - 3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$$

That is, $$8\mu = 2 \Rightarrow \mu = {1 \over 4}$$