# a) Find all vectors v such that (1, 2, 1) \times v = (3, 1, -5). Describe the shape these vectors...

## Question:

a) Find all vectors {eq}v {/eq} such that {eq}(1, 2, 1) \times v = (3, 1, -5) {/eq}. Describe the shape these vectors form.

b) Explain why there is no vector {eq}v {/eq} such that {eq}(1, 2, 1) \times v = (3, 1, 5) {/eq}.

## Operations of Vectors

The cross product of two vectors is a vector orthogonal on both vectors involved in the operation and defined as

{eq}\displaystyle \langle a_1,b_1,c_1 \rangle \times \langle a_2,b_2,c_2\rangle= \begin{array}{|ccc|} \mathbf{i} &\mathbf{j}&\mathbf{k}\\ a_1&b_1&c_1\\ a_2&b_2&c_2 \end{array} =\langle b_1c_2-b_2c_1, -a_1c_2+a_2c_1, a_1b_2-a_2b_1\rangle. {/eq}

To check if two vectors are orthogonal, we can check if their dot product is zero,

{eq}\displaystyle \langle a_1,b_1,c_1\rangle \cdot \langle a_2,b_2,c_2\rangle =a_1a_2+b_1b_2+c_1c_2\\ \displaystyle \text{ or }\vec{u} \cdot \vec{v} =|\vec{u}||\vec{v}|\cos \theta, \text{ where }\theta \text{ is the angle between the two vectors}. {/eq}

a) To find the vector {eq}\displaystyle \vec{v}=\langle a, b, c \rangle {/eq} such that {eq}\displaystyle \langle 1,2,1\rangle \times \langle a,b,c\rangle=\langle 3,1,-5\rangle {/eq}

we will use the cross product rule, as below.

{eq}\displaystyle \begin{align}\langle 1,2,1\rangle \times \langle a,b,c\rangle&=\langle 3,1,-5\rangle\\ \left|\begin{array}{ccc} 1 &2&1\\ a&b&c \end{array}\right|&=\langle 3,1,-5\rangle\\ \langle 2c-b, -c+a, b-2a\rangle&=\langle 3,1,-5\rangle\\ \implies &\begin{cases}&-b&+2c&=3\\ a& &-c&=1\\ -2a&+b& & =-5 \end{cases}\iff &\begin{cases} a& &-c&=1\\ &-b&+2c&=3\\ -2a&+b& & =-3 \end{cases}\iff &\begin{cases} a& &-c&=1\\ &-b&+2c&=3\\ &b&-2c & =-3 \end{cases}\iff \begin{cases} a& &-c&=1\\ &-b&+2c&=3\\ && 0& =0 \end{cases}\\ \implies &b=-3+2c \text{ and }a=1+c \text{ for any }c\\ \implies &\boxed{\vec{v}=\langle 1+c, -3+2c, c \rangle, c\in\mathbb{R}}. \end{align} {/eq}

So, we obtain a set of vectors {eq}\displaystyle \vec{v} {/eq} that are perpendicular on the two vectors, {eq}\displaystyle \langle 1,2,1\rangle, \langle 3,1,-5\rangle, {/eq}

therefore, the set of vectors are in a plane perpendicular on the plane determined by the two vectors.

b) There is no vector {eq}\displaystyle \vec{v} {/eq} such that {eq}\displaystyle \langle 1,2,1\rangle \times \langle a,b,c\rangle=\langle 3,1,5\rangle {/eq}

because the cross product is perpendicular on the two vectors in the cross product

and {eq}\displaystyle \langle 3,1,5\rangle, \langle 1,2,1\rangle {/eq} are not orthogonal, since their dot product is nonzero {eq}\displaystyle \langle 3,1,5\rangle \cdot \langle 1,2,1\rangle=3+2+5\neq 0. {/eq} 