Question

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

• A. \frac{-16}{13}
• B. \frac{16}{13}
• C. \frac{-13}{16}
• D. \frac{13}{16}

Answer 1

The correct value of $\mu$ is $\frac{13}{10}$ (none of the given options).

Answer 2

Solution:

1. Coordinates of Q:
   The line is given by

   $$\vec{r} = (1, -1, 2) + \mu(-3, 1, 5).$$

   Hence,

   $$Q=(1-3\mu, -1+\mu, 2+5\mu).$$

2. Vector PQ:
   Point $P=(2, 1, 5)$, so

   $$\overrightarrow{PQ}= Q-P = \bigl((1-3\mu)-2, (-1+\mu)-1, (2+5\mu)-5\bigr) = (-1-3\mu, -2+\mu, -3+5\mu).$$

3. Condition for Parallelism:
   A vector is parallel to a plane if it is perpendicular to the plane’s normal.
   The plane is

   $$3x-y+4z=1,$$

   so its normal is

   $$\mathbf{n}=(3,-1,4).$$

   Therefore,

   $$\overrightarrow{PQ}\cdot \mathbf{n}=0.$$

4. Dot Product Calculation:

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

   Expanding:

   $$-3-9\mu +2-\mu -12+20\mu= 0.$$

   Combine like terms:

   $$(-3+2-12) + (-9\mu-\mu+20\mu)= -13 + 10\mu= 0.$$

   Thus,

   $$10\mu=13\quad\Longrightarrow\quad \mu=\frac{13}{10}.$$

5. Observing the Options:
   The computed answer $\frac{13}{10}$ does not match any of the provided options (A: $-\frac{16}{13}$, B: $\frac{16}{13}$, C: $-\frac{13}{16}$, D: $\frac{13}{16}$). Hence, the correct value is $\frac{13}{10}$.

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Minimal Explanation:
Express $Q$ in terms of $\mu$; compute $\overrightarrow{PQ}=Q-P$; require $\overrightarrow{PQ}$ to be perpendicular to the normal $(3, -1, 4)$; solve $(-1-3\mu)3+(-2+\mu)(-1)+(-3+5\mu)4=0$ to get $\mu=\frac{13}{10}$.