Question

find the scalar component of b in the direction of a and the vector projection of b onto a, where a = <-1,1,2> and b = <-1,7,4>. 1. scal_ab = 2. proj_ab = answers # your answer score 01 - -/1 02 - -/1

Explanation:

Step1: Recall the formula for scalar component

The scalar component of $\mathbf{b}$ in the direction of $\mathbf{a}$ is given by $\text{scal}_{\mathbf{a}}\mathbf{b}=\frac{\mathbf{a}\cdot\mathbf{b}}{\vert\mathbf{a}\vert}$. First, find the dot - product $\mathbf{a}\cdot\mathbf{b}$ and the magnitude of $\mathbf{a}$, $\vert\mathbf{a}\vert$.
The dot - product of two vectors $\mathbf{a}=\langle a_1,a_2,a_3
angle$ and $\mathbf{b}=\langle b_1,b_2,b_3
angle$ is $\mathbf{a}\cdot\mathbf{b}=a_1b_1 + a_2b_2+a_3b_3$. Here, $\mathbf{a}=\langle - 1,1,2
angle$ and $\mathbf{b}=\langle - 1,7,4
angle$, so $\mathbf{a}\cdot\mathbf{b}=(-1)\times(-1)+1\times7 + 2\times4=1 + 7+8 = 16$.
The magnitude of $\mathbf{a}$ is $\vert\mathbf{a}\vert=\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}=\sqrt{(-1)^{2}+1^{2}+2^{2}}=\sqrt{1 + 1+4}=\sqrt{6}$.

Step2: Calculate the scalar component

Then, $\text{scal}_{\mathbf{a}}\mathbf{b}=\frac{\mathbf{a}\cdot\mathbf{b}}{\vert\mathbf{a}\vert}=\frac{16}{\sqrt{6}}=\frac{8\sqrt{6}}{3}$.

Step3: Recall the formula for vector projection

The vector projection of $\mathbf{b}$ onto $\mathbf{a}$ is given by $\text{proj}_{\mathbf{a}}\mathbf{b}=\frac{\mathbf{a}\cdot\mathbf{b}}{\vert\mathbf{a}\vert^{2}}\mathbf{a}$.
We know that $\mathbf{a}\cdot\mathbf{b}=16$ and $\vert\mathbf{a}\vert^{2}=(-1)^{2}+1^{2}+2^{2}=6$, and $\mathbf{a}=\langle - 1,1,2
angle$.
So $\text{proj}_{\mathbf{a}}\mathbf{b}=\frac{16}{6}\langle - 1,1,2
angle=\langle-\frac{8}{3},\frac{8}{3},\frac{16}{3}
angle$.

Answer:

1. $\frac{8\sqrt{6}}{3}$
2. $\langle-\frac{8}{3},\frac{8}{3},\frac{16}{3}

angle$