Solve the problem.
-Find a vector of magnitude 9 in the direction opposite to the direction of $\mathbf { v } = \frac { 1 } { 3 } \mathbf { i } + \frac { 1 } { 3 } \mathbf { j } - \frac { 1 } { 3 } \mathbf { k }$.
A) $9 \sqrt { 3 } \left( - \frac { 1 } { 3 } i - \frac { 1 } { 3 } j + \frac { 1 } { 3 } k \right)$
B) $9 \sqrt { 3 } \left( \frac { 1 } { 3 } \mathbf { i } + \frac { 1 } { 3 } \mathbf { j } - \frac { 1 } { 3 } \mathbf { k } \right)$
C) $9 \left( \frac { 1 } { 3 } \mathbf { i } + \frac { 1 } { 3 } \mathrm { j } - \frac { 1 } { 3 } \mathbf { k } \right)$
D) $9 \left( - \frac { 1 } { 3 } \mathbf { i } - \frac { 1 } { 3 } \mathbf { j } + \frac { 1 } { 3 } \mathbf { k } \right)$

Question

Quizplus · Accepted Answer

The correct answer is A because the vector $9 \sqrt { 3 } \left( - \frac { 1 } { 3 } i - \frac { 1 } { 3 } j + \frac { 1 } { 3 } k \right)$ has magnitude 9 and is in the opposite direction to the vector $\mathbf { v } = \frac { 1 } { 3 } \mathbf { i } + \frac { 1 } { 3 } \mathbf { j } - \frac { 1 } { 3 } \mathbf { k }$.

Solve the Problem v=13i+13j−13k\mathbf { v } = \frac { 1 } { 3 } \mathbf { i } + \frac { 1 } { 3 } \mathbf { j } - \frac { 1 } { 3 } \mathbf { k }v=31​i+31​j−31​k

Solve the Problem $\mathbf { v } = \frac { 1 } { 3 } \mathbf { i } + \frac { 1 } { 3 } \mathbf { j } - \frac { 1 } { 3 } \mathbf { k }$