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Find the components of all vectors v which have length 2 and are perpendicular to both the lines $x = 4 + 3t, y = 2 − t, z = 1 + 5t$ and $x − y + z = 2$, $3x + 2y − 4z = 6$.

So far I know that the vector $(1, -1, 1)$ is normal to the line $x-y+z=2$, and so is $<3, 2, -4>$ to the line $3x+2y-4z=6$. So I took their cross products, and that gives me the vector perpendicular to those lines.

For the line given in parametric form, I'm starting to get really confused - don't I have to substitute those into one of the equations i.e $(4+3t) - (2-t) + (1+5t) = 2$? I'm doing that and whatever value I'm getting for t does not satisfy the equation. I'm not sure if I'm in the right track or not. :(

1 Answers 1

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I am assuming that your second line is the intersection of the two planes $x-y+z=2$ and $3x+2y-4z=6$ namely $(2+2t,7t,5t)$.

What remains is then to find the vectors perpendicular to $(4+3t,2-t,1+5t)$ and $(2+2t,7t,5t)$. To do this take the cross product of the direction vectors for each line and get $v=(-40,-5,23)$.

The question asks for all vectors with length 2 so we get $w=\pm\frac{2v}{||v||}=\pm\frac{2}{\sqrt{2154}}\begin{pmatrix}-40\\-5\\23\end{pmatrix}$