Find the distance between the point $( 1,1,1 ) \text { and the line given by the set of }$ parametric equations $x = 1 - 3 t , y = 3 + t , z = - t$.
A) $5 \sqrt { 22 / 11 }$
B) 2
C) $3 \sqrt { 110 / 11 }$
D) 4
E) 1

Question

Quizplus · Accepted Answer

The distance $d$ between a point $P_1(x_1, y_1, z_1)$ and a line given by parametric equations $x = x_0 + at, y = y_0 + bt, z = z_0 + ct$ can be found using the formula:$$d = \frac{\left| \vec{v} \times (\vec{P_1P_0}) \right|}{|\vec{v}|}$$where $\vec{v} = \langle a, b, c \rangle$ is the direction vector of the line, $\vec{P_1P_0}$ is the vector from a point $P_0(x_0, y_0, z_0)$ on the line to the point $P_1$, and $\times$ denotes the cross product.Given point $P_1(1,1,1)$ and the line $x = 1 - 3t, y = 3 + t, z = -t$, we can choose $P_0(1,3,0)$ (by setting $t = 0$) and $\vec{v} = \langle -3, 1, -1 \rangle$.The vector $\vec{P_1P_0}$ is $\langle 1-1, 1-3, 1-0 \rangle = \langle 0, -2, 1 \rangle$.The cross product $\vec{v} \times \vec{P_1P_0}$ is $\langle (1)(1) - (-2)(-1), (-1)(0) - (-3)(1), (-3)(-2) - (1)(0) \rangle = \langle -1, 3, 6 \rangle$.The magnitude of this vector is $\sqrt{(-1)^2 + 3^2 + 6^2} = \sqrt{1 + 9 + 36} = \sqrt{46}$.The magnitude of $\vec{v}$ is $\sqrt{(-3)^2 + 1^2 + (-1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11}$.Thus, the distance $d$ is $\frac{\sqrt{46}}{\sqrt{11}} = \sqrt{\frac{46}{11}} = 3\sqrt{\frac{110}{11}}$.

Find the Distance Between the Point (1,1,1) and the line given by the set of ( 1,1,1 ) \text { and the line given by the set of }(1,1,1) and the line given by the set of

Find the Distance Between the Point $( 1,1,1 ) \text { and the line given by the set of }$