Question

The equation of a line in cartesian form passing through (0, 0, 0) and (4, 3, c) and parallel to \(\overset{⃗}{a}\times \overset{⃗}{b}\) where \(\overset{⃗}{a} = 2\hat{i}+\hat{j}+2\hat{k}\), \(\overset{⃗}{b} = 3\hat{i}-4\hat{j}\) is

• A. \(\frac{x-4}{8} = \frac{y-3}{6} = \frac{z+\frac{11}{2}}{11}\)
• B. \(\frac{x-4}{8} = \frac{y+3}{6} = \frac{z+\frac{11}{2}}{-11}\)
• C. \(\frac{x-4}{8} = \frac{y-3}{6} = \frac{z+\frac{11}{2}}{-11}\)
• D. \(\frac{x+4}{8} = \frac{y-3}{6} = \frac{z+\frac{11}{2}}{-11}\)

Hint:

Compute a cross b, then find c so that (4,3,c) lies along it.

Answer 1

The Correct Option is C

Step 1: Cross product
Components: $i: 1\cdot0 - 2\cdot(-4) = 8$; $j: -(2\cdot0 - 2\cdot3) = 6$; $k: 2(-4) - 1\cdot3 = -11$.

Step 2: Collinearity
The origin, $(4,3,c)$ and the direction $(8,6,-11)$ are on one line, so $\frac{4}{8} = \frac{3}{6} = \frac{c}{-11}$ giving $c = -\frac{11}{2}$.

Step 3: Equation
Using $(4,3,-\frac{11}2)$ as the point gives option (C).

Final Answer:
Option C. \[ \boxed{\text{(C)}\ \frac{x-4}{8}=\frac{y-3}{6}=\frac{z+\frac{11}{2}}{-11}} \]