Question

The cartesian equation of the line passing through the point \(\hat{i}+2\hat{j}+3\hat{k}\) and perpendicular to the vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(2\hat{i}-\hat{j}+\hat{k}\) is

• A. \(\frac{x+1}{2}=y+2=\frac{z+3}{3}\)
• B. \(\frac{x-1}{2}=y-2=\frac{z-3}{3}\)
• C. \(\frac{x-1}{2}=y-2=\frac{z-3}{-3}\)
• D. \(\frac{x+1}{2}=y+2=\frac{z+3}{-3}\)

Hint:

The direction of the line is the cross product of the two given vectors.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A line perpendicular to two vectors is parallel to their cross product.

Step 2: Cross product
\(\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\2&-1&1\end{vmatrix}=\hat{i}(1+1)-\hat{j}(1-2)+\hat{k}(-1-2)=2\hat{i}+\hat{j}-3\hat{k}\).

Step 3: Write the line
The line passes through \((1,2,3)\) with direction ratios \(2,1,-3\).

Step 4: Cartesian form
\(\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{-3}\).

Final Answer:
The line is (x-1)/2 = y-2 = (z-3)/(-3). \[ \boxed{\text{(C) }\dfrac{x-1}{2}=y-2=\dfrac{z-3}{-3}} \]