A line has Cartesian equations x−p = (y+2)/q = 3−z and a plane has equation r ∙ [1,−1,−2] = −3. In the case where the angle θ between the line and the plane satisfies sin⁡θ=1/√6 and the line intersects the plane at z = 0. Find p and q.

This question gives hints as to how you should solve it. Because it tells you the angle between the line and the plane, this should indicate that doing something to do with finding the angle between the line and the plane, will hep find either p, q or both. Similarly, they give you one of the coordinates to the point of intersection, so this should indicate, something to do with finding the intersection point between the line and the plane should also help to find either p, q or both.We can't directly find the angle between the plane and the line, but we can fnd the angle betweeen the line and the normal to the plane (which can be used to find the angle we want). So we need the direction vector of the line, and the normal to the plane. We know the normal to the plane is [1,−1,−2], from looking at the equation given to us. To work out the direction vector for the line, we must make the three Cartesian equations for the line equal to some parameter/variable, s, and express it in vector form.x−p = (y+2)/q = 3−z = s=> x = s + p=> y = qs - 2=> z = 3 - sNow this can be expressed in vector form:=> (x, y, z) = (1, q, -1)s + (p, -2, 3)From this you can see the direction vector is (1, q, -1) as it is multiplied by the parameter, s. To find the angle between this direction vector and the normal to the plane, we can use the dot product of these two vectors.d = (1, q, -1), n = (1,−1,−2)d ∙ n = 1 - q + 2d ∙ n = |d| |n| cos x = √(1+q2+1) √(1+1+4) cos x = √(2+q2) √(6) cos xNote that x is not the angle between the plane and the line, but the angle betweeen the normal and the line, so x = 90 - θ. However, we can introduce theta into this, recalling that cos (90 - θ) = sin θ.=> 3 - q = √(2+q2) √(6) sin x = √(2+q2) √(6) 1/√(6) = √(2+q2)Now rearrange to find q:3 - q = √(2+q2)=> 9 - 6q + q2 = 2 + q2=> 6q = 7=> q = 7/6Now work on finding the intersection point between the line and the plane. Start by substituing in the equation for the line into the caresian equation for the plane. From the equation for the plane given to us, and performing the dot product, the cartesian equation is:x - y - 2z = -3Now substitue in the line. Note, we know z is zero, so we can ignore that term:s + p - qs + 2 = -3We know q:s + p - (7/6)s + 2 = -3We can also work out s, because we know z = 0 at the point of intersection. From the equation for z interms of s at the beginning:z = 3 - s => 0 = 3 - s => s = 3So now we have:3 + p - (7/6)3 + 2 = -3 = > p = -8 + 7/2 => p = -9/2Answer: p = -9/2, q = 7/6