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: Direction
\(\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\\2&1&2\\3&-4&0\end{vmatrix} = (0+8)\hat i - (0-6)\hat j + (-8-3)\hat k = 8\hat i+6\hat j-11\hat k\).

Step 2: Find c
The line passes through the origin with direction \((8,6,-11)\), so \((4,3,c) = \frac12(8,6,-11)\). Thus \(c = -\frac{11}{2}\).

Step 3: Write the line
Through \((4,3,-\frac{11}{2})\) with direction \((8,6,-11)\):
\[ \frac{x-4}{8} = \frac{y-3}{6} = \frac{z+\frac{11}{2}}{-11} \]

Step 4: Result
Option (C). Option (A) has the wrong sign of the \(z\) direction, and (B) and (D) use the wrong points.

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