Chapter 9.4, Problem 1E

A vector in three dimensions can be written in either of two forms: in coordinate form as v = 〈a₁, a₂, a₃〉 and in terms of the __________ vectors i, j, and k as v = _______________. The magnitude of the vector v is | v | = _______________. So $\langle 4,-2,4\rangle =\boxed{}i+\boxed{}j+\boxed{}k$ and $7j-24k=\langle \boxed{},\boxed{},\boxed{}\rangle$.

To determine

To fill: The vector $v\langle a_1,a_2,a_3\rangle$, $\langle 4,-2,4\rangle$ in the terms of unit vector i, j and k and find the magnitude of the vector $v\langle a_1,a_2,a_3\rangle$, $\langle 4,-2,4\rangle$.

Answer to Problem 1E

The vector $v\langle a_1,a_2,a_3\rangle$ in the terms of unit vector i, j and k is $a_{1}i+a_{2}j+a_{3}k$ and magnitude of the vector is $\left|v\right|=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2}$.

The vector $\langle 4,-2,4\rangle$ in the terms of unit vector i, j and k is $4i-2j+4k$ and magnitude of the vector is $\left|4,-2,4\right|=6$.

Explanation of Solution

Given:

The vector v are $\langle a_1,a_2,a_3\rangle$ and $\langle 4,-2,4\rangle$.

Calculation:

Section (1):

Formula used:

The vector $v=\langle a_1,a_2,a_3\rangle$ can be written in terms of unit vector i, j and k as,

$v\langle a_1,a_2,a_3\rangle =a_{1}i+a_{2}j+a_{3}k$

And the magnitude of the vector $v\langle a_1,a_2,a_3\rangle$ is $\left|v\right|=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2}$.

Hence, the vector $v\langle a_1,a_2,a_3\rangle$ in the terms of unit vector i, j and k is $a_{1}i+a_{2}j+a_{3}k$ and magnitude of the vector is $\left|v\right|=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2}$.

Section (2):

From section (1),

Vector $\langle 4,-2,4\rangle$ can be written in the terms of unit vector i, j and k as,

$\begin{array}{c}\langle 4,-2,4\rangle =4i+\left(-2j\right)+4k\\ =4i-2j+4k\end{array}$

And the magnitude of the vector $\langle 4,-2,4\rangle$ is

$\begin{array}{c}\left|4,-2,4\right|=\sqrt{{\left(4\right)}^2+{\left(-2\right)}^2+{\left(4\right)}^2}\\ =\sqrt{16+4+16}\\ =\sqrt{36}\\ =6\end{array}$

Hence, the vector $\langle 4,-2,4\rangle$ in the terms of unit vector i, j and k is $4i-2j+4k$ and magnitude of the vector is $\left|4,-2,4\right|=6$.