# Find b and c so that (7, b, c) is orthogonal to both (1, 2, 9) and (1, -2, 1).

## Question:

Find {eq}b {/eq} and {eq}c {/eq} so that {eq}(7, b, c) {/eq} is orthogonal to both {eq}(1, 2, 9) {/eq} and {eq}(1, -2, 1). {/eq}

## Cross Product

Let {eq}\vec{u}=(u_1,u_2,u_3) {/eq} and {eq}\vec{v} = (v_1,v_2,v_3) {/eq} be nontrivial vectors in {eq}\mathbb{R}^3. {/eq} The cross product is new function that determines a vector that is normal to both vectors and is defined as: {eq}\vec{w} = \vec{u} \times \vec{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1 v_3, u_1 v_2 - u_2 v_1). {/eq}

## Answer and Explanation:

Consider the vectors {eq}\vec{u} = (1, 2, 9) {/eq} and {eq}\vec{v} = (1, -2, 1). {/eq} Now the cross product of these vectors determines a vector that is orthogonal to both of these vectors. Namely,

{eq}\vec{w} = \vec{u}\times \vec{v} = (20, 8, -4). {/eq}

Now scaling this vector so that the first component is length 7 does not influence the fact the vector is orthogonal to both {eq}\vec{u} {/eq} and {eq}\vec{v}. {/eq} Hence:

{eq}\vec{z} = \dfrac{7}{20} \vec{w} = (7, 14/5, -7/5) {/eq}

Subsequrently, we let {eq}b = \dfrac{14}{5} {/eq} and {eq}c = -\dfrac75. {/eq}

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