Question

The distance between the line \(\overline{\mathrm{r}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})\) and the plane \(\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=4\) is

• A. \(\frac{1}{\sqrt{6}}\) units
• B. \(\frac{3}{\sqrt{6}}\) units
• C. \(\frac{2}{\sqrt{6}}\) units
• D. \(\frac{5}{\sqrt{6}}\) units

Hint:

Check that the line is parallel to the plane, then find the distance of any point of the line from the plane.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
If the direction of the line is perpendicular to the normal of the plane, the line is parallel to the plane and its distance equals the distance of any of its points from the plane.

Step 2: Check Parallelism
Direction \(=\hat i-2\hat j\), normal \(=2\hat i+\hat j+\hat k\).
Dot product \(=2-2+0=0\), so the line is parallel to the plane.

Step 3: Calculation
Take the point \((3,-2,1)\) on the line. The plane is \(2x+y+z-4=0\).
Distance \(=\dfrac{|6-2+1-4|}{\sqrt{4+1+1}}=\dfrac{1}{\sqrt6}\)

Step 4: Compare the options
The distance \(\dfrac{1}{\sqrt6}\) units matches the first option.

Final Answer:
The line is 1 over root 6 units from the plane. \[ \boxed{\text{(A) }\dfrac{1}{\sqrt{6}}\text{ units}} \]