alleryThe cartesian equation of a line through $(x_1,y_1,z_1)$ with direction ratios $a,b,c$ is:
A$\tfrac{x-x_1}{a}+\tfrac{y-y_1}{b}+\tfrac{z-z_1}{c}=1$
B$\tfrac{a}{x-x_1}=\tfrac{b}{y-y_1}=\tfrac{c}{z-z_1}$
C$\tfrac{x-x_1}{a}=\tfrac{y-y_1}{b}=\tfrac{z-z_1}{c}$
D$ax+by+cz=0$
Answer & Solution
Correct answer: C. $\tfrac{x-x_1}{a}=\tfrac{y-y_1}{b}=\tfrac{z-z_1}{c}$
The symmetric (cartesian) form of a line is (x−x₁)/a = (y−y₁)/b = (z−z₁)/c.
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: