alleryPlane passing through origin and perpendicular to line (x-1)/2 = (y-2)/3 = (z-3)/-1:
A2x + 3y - z = 0 (normal = line direction)
Bx + y + z = 0
Cx - y - z = 0
Dz = 0
Answer & Solution
Correct answer: A. 2x + 3y - z = 0 (normal = line direction)
Direction of line = (2, 3, -1).
Plane perpendicular to line has this as normal.
**Through origin:** 2x + 3y - z = 0.
Keep practising Unit 8A: Areas Between Curves
Browse all Unit 8A: Areas Between Curves questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
Let $\mathbf{A}=\langle 1,2,3\rangle$, $\mathbf{B}=\langle 2,0,1\rangle$, and $\mathbf{C}=For nonzero vector $\mathbf{A}$, which expression gives a vector in the same direction as When subtracting vectors, $\vec A-\vec B$ is evaluated asThe projection of vector $\vec A$ on vector $\vec B$ is given byWhich statement about a null vector is correct?If $\omega$ is a non-real cube root of unity, find the value of $1 + \omega + \omega^2$.Find the modulus of the complex number $z = 3 + 4i$.The locus of the point $z$ in the Argand plane satisfying $|z - 2| = |z + 2|$ is: