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Chapter 11: Problem 37

Find a vector orthogonal to the given vectors. $$\langle 8,0,4\rangle \text { and }\langle-8,2,1\rangle$$

Short Answer

Expert verified
Question: Find the vector orthogonal to the given vectors $$\vec{A} = \langle 8,0,4 \rangle$$ and $$\vec{B} = \langle -8,2,1 \rangle$$. Answer: The vector orthogonal to both given vectors is $$\langle -8, -40, 16 \rangle$$.

Step by step solution

01

Recall the cross product formula

To find the cross product of two vectors $$\vec{A} = \langle a_1, a_2, a_3 \rangle$$ and $$\vec{B} = \langle b_1, b_2, b_3 \rangle$$, we'll use the following formula: $$\vec{A} \times \vec{B} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle$$
02

Plug in the vectors' components into the formula

We are given the vectors $$\vec{A} = \langle 8,0,4 \rangle$$ and $$\vec{B} = \langle -8,2,1 \rangle$$. Plugging their components into the cross product formula from Step 1: $$\vec{A} \times \vec{B} = \langle 0\cdot1 - 4\cdot2, 4\cdot(-8) - 8\cdot1, 8\cdot2 - 0\cdot(-8) \rangle$$
03

Compute the cross product

Now, let's compute the cross product: $$\vec{A} \times \vec{B} = \langle -8, -32-8, 16 \rangle$$
04

Simplify the resulting vector

Simplifying the resulting vector, we get: $$\vec{A} \times \vec{B} = \langle -8, -40, 16 \rangle$$ Thus, the vector orthogonal to both given vectors is $$\langle -8, -40, 16 \rangle$$.

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Most popular questions from this chapter

An ant walks due east at a constant speed of \(2 \mathrm{mi} / \mathrm{hr}\) on a sheet of paper that rests on a table. Suddenly, the sheet of paper starts moving southeast at \(\sqrt{2} \mathrm{mi} / \mathrm{hr} .\) Describe the motion of the ant relative to the table.

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