If the lines ${{x - k} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$ and
${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$ are co-planar, then the value of k is _____________.
Answer (integer)
1
Solution
$$\left| {\matrix{
{k + 1} & 4 & 6 \cr
1 & 2 & 3 \cr
3 & 2 & 1 \cr
} } \right| = 0$$<br><br>$\Rightarrow$ $(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$<br><br>$\Rightarrow$ $k = 1$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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