JEE Mains · Maths · STD 12 - 11. three dimension geometry
The line, that is coplanar to the line \(\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\), is
- A \(\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-5}{5}\)
- B \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\)
- C \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}\)
- D \(\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\)
Answer & Solution
Correct Answer
(B) \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\)
Step-by-step Solution
Detailed explanation
Condition of co-planarity \(\left|\begin{array}{lll}x_2-x_1 & a_1 & a_2 \\ y_2-y_1 & b_1 & b_2 \\ z_2-z_1 & c_1 & c_2\end{array}\right|=0\) Where \(a_1, b_1, c_1\) are direction cosine of \(1^{\text {st }}\) line and \(a_2, b_2, c_2\) are direction cosine of \(2^{\text {nd }}\)…
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