Find the projection of the straight line $(x-1)/2 = (y-1)/3 = (z+1)/-1$ on the plane $x+y+2z = 6$.

Question

I have got the answer, seeking intuition from https://www.doubtnut.com/question-answer/find-the-equations-of-the-projection-of-the-line-x-1-2y-1-3z-2-4-on-the-plane-2x-y-4z1-1116852

The answer is $\frac{x-5}{3} = \frac{y-7}{5} = \frac{z+3}{-4}$

Why didn't my initial approach work ?

Projection of a line on a plane is line (if original line was not normal to it) Let its dc's : $l,m,n$. It will lie in plane, so

$l + m + 2n = 0 \: ...(1)$

$\cos \theta = \frac{2l + 3m -n}{\sqrt{14}} \ ...(2)$, $\theta$ is angle between original line and plane

$\sin \theta = \frac{1}{\sqrt{14 \cdot 6}} (2.1 + 3.1 + (-1).2) \ ...(3)$

EDIT:

$ \sin(\theta) = \frac{\sqrt{3/2}}{\sqrt{14}} $

$ \cos(\theta) = \frac{5 / \sqrt2}{\sqrt{14}} = \frac {2l + 3m -n}{\sqrt{14}}$ ... from (2)

$ \implies 2l + 3m -n = \frac{5}{\sqrt2} $ ..(4)

how can I find ratio of $l,m,n$ from (3) and (4)?

(Since dc's, so can also use $l^2 + m^2 + n^2 = 1$ - it's a post thought, but no help)

Answer 1

Finding direction vector of the line using the method you started with is possible. But we need additional steps to find a point on the line. But to answer your question, here is how we can find direction vector using the method you started with.

Say unit direction vector of the line is $ \ \hat d = (l, m, n)$.

As the line is in the plane $x + y + 2z = 6$, dot product of its direction vector with the normal vector to the plane must be zero.

$l + m + 2n = 0 \implies m + 2n = -l\tag1$

If angle between original line and the normal vector to the plane is $\theta, \ $ then

$(1 \cdot 2 + 1 \cdot 3 + 2 \cdot -1) = \sqrt6 \cdot \sqrt{14} \cos\theta \implies \displaystyle \cos\theta = \sqrt{\frac{3}{28}}, \ \sin \theta = \frac{5}{\sqrt{28}}$

As the angle between the original line and its projection onto the plane is $(90^0 - \theta)$,

$\displaystyle 2l + 3m - n = \sqrt{14} \cdot \frac{5}{\sqrt{28}} = \frac{5}{\sqrt2} \implies 3m-n = \frac{5}{\sqrt2} - 2l\tag2$

Solving $(1)$ and $(2)$,

$\displaystyle m = \frac{5 (\sqrt2-l)}{7} \land n = - \frac{5 + l \sqrt2}{7\sqrt2}$

As $(l. m. n)$ is unit vector, $l^2+m^2+n^2 = 1 \implies l^2 + (\frac{5 (\sqrt2-l)}{7})^2 + (- \frac{5 + l \sqrt2}{7\sqrt2})^2 = 1$

This simplifies to, $ \ \displaystyle l^2 - \frac{3 \sqrt2 l}{5} + \frac{9}{50} = 0 \implies (l - \frac{3}{5\sqrt2})^2 = 0$

So we get, $l = \frac{3}{5\sqrt2} \land m = \frac{5}{5\sqrt2} \land n = - \frac{4}{5 \sqrt2} $

That gives direction of line as $(3, 5, -4)$.

Answer 2

A straightforward method is:

1. Take the normal of the plane : $\vec{i}$ + $\vec{j}$ + 2.$\vec{k}$ ( A line with direction cosines <1, 1, 2>)
2. Take two points on the given line; for example A: <1, 1, -1>, B :<3, 4, -2>
3. Solve for line with direction cosines <1, 1, 2>, passing through A and B and find their intersection with the plane. This will give you 2 points on the plane that are projections of A and B. Let them be A'($A'_x, A'_y, A'_z$) and B'($B'_x, B'_y, B'_z$)
4. The projection of the line on the plane is $\frac{x - A'_x}{A'_x - B'_x}$ = $\frac{y - A'_y}{A'_y - B'_y}$ = $\frac{z - A'_z}{A'_z - B'_z}$

Hope this helps