Question

The position vector of the point at which the line joining the points \(\hat{i}+2\hat{j}-\hat{k}\) and \(3\hat{i}+4\hat{j}-2\hat{k}\) cuts the plane \(\vec{r}\cdot (\hat{j}-\hat{k})=-1\) is

• A. \(-\frac{5}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}\)
• B. \(-\frac{5}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\)
• C. \(\frac{5}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}\)
• D. \(\frac{5}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\)

Hint:

Write the line in parametric vector form and substitute into the plane equation to find the parameter.

Answer 1

The Correct Option is A

Step 1: Test the Plane Equation
The plane is $y-z=-1$. Check which option satisfies $y-z=-1$.

Step 2: Evaluate
Option (A): $-\dfrac{2}{3}-\dfrac{1}{3}=-1$, satisfied. Option (B): $-\dfrac{2}{3}+\dfrac{1}{3}=-\dfrac{1}{3}$, no. Option (C): $-\dfrac{2}{3}-\dfrac{1}{3}=-1$, also satisfied, so check the line.

Step 3: Check the Line
The direction is $(2,2,-1)$. From the first point $(1,2,-1)$, option (A) gives displacement $\left(-\dfrac{8}{3},-\dfrac{8}{3},\dfrac{4}{3}\right)=-\dfrac{4}{3}(2,2,-1)$, so it lies on the line. Option (C) does not (its x displacement is positive but y is negative).

Final Answer:
Only option (A) lies on both the line and the plane. \[ \boxed{\text{(A) }-\dfrac{5}{3}\hat{i}-\dfrac{2}{3}\hat{j}+\dfrac{1}{3}\hat{k}} \]