#86 part \'a\' Intersection if and only if v (a, b, c,) = 0 explanation of the r

Question

#86 part 'a'

Intersection if and only if v (a, b, c,) = 0 explanation of the result. Tilted ellipse Consider the curve r(t) - (cos 1, sin 1, c sin t) for o 1 2r, where c is a real number. What is the equation of the plane P in which the curve lies? What is the angle between P and the xy-plane? prove that the curve is an ellipse on P. Distance from a point to a plane Show that the point in the plane ax + by + cz = d nearest the origin is p(ad/D2, bd./D2,cd/d2), where D2 = a2 + b2 + c2. Conclude that the least distance from the plane to the origin is d /D. (Hints: The least distance is along a normal to the plane) Show that the least distance from the point p0(x0,-y0,z0) to the plane ax + by + cz = d is [ax0 + by0 + cz0 - d]/D. (Hints; Find the point p on the plane closest to p0.) Projections Find the projection of the position vector (2, 3, -4) on the plane 3x + 2y + -z = 0. Ellipsoid plane intersection Let E be the ellipsoid x2/9 + y2/4 + z = 1, p be the plane z = Ax = By, and C be

Explanation / Answer

First we note that any line that is perpendicular to ax+by+cz=d is also perp to ax+by+cz=0, since these planes are parallel to each other. Let's find two vector in the plane ax+by+cz=0. By inspection, v1=(b,-a, 0) and v2=(0, c, -b) are two such vectors. Now let's find the vector perpendicular to the plane of these two. That vector is v1Xv2, and writing out the determinant for the cross-product, we get the vector (ab, b2,bc) which is just the vector v3=(a,b,c) scaled by a factor b.

So we want to find the line passing through the origin that lies along this vector. Clearly {y/x = b/a; z/x = c/a} does the trick. Now let's find where this line, which is perpendicular to the given plane and passes through the origin, intersects the plane (ax+by+cz)=d. Substitute y=x(b/a), z=x(c/a) (from the Eq. of the straight line) to get x(a+b2/a+c2/a)=d, giving x=ad/(a2+b2+c2). Also, y=(b/a)x =bd/(a2+b2+c2) and z=(c/a)x = =cd/(a2+b2+c2). This proves the first assertion. The second assertion is easily proven from the first. Calculate (x-0)2 + (y-0)2 + (z-0)2 = distance between the point (0,0,0) and the point (x,y,z) = |d|/(a2+b2+c2) where the | | is because d2 = |d|.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.