What is the cross product of $< 4, 5, 0 >$ $<4 , 5 ,0 >$ and $< 4, 1, - 2 >$ $<4, 1 ,-2 >$ ?

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What is the cross product of $< 4, 5, 0 >$ $<4 , 5 ,0 >$ and $< 4, 1, - 2 >$ $<4, 1 ,-2 >$ ?

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Answer 1 · 2017-05-24T16:20:12

$< - 10, - 8, - 16 >$ $< -10, -8, -16>$

Explanation:

We'll call the vector $< 4, 5, 0 > \to A$ $< 4, 5, 0 > vec A$ , and the vector $< 4, 1, - 2 > \to B$ $< 4, 1, -2> vec B$

The cross product of $\to A$ $vec A$ and $\to B$ $vec B$ is a vector $\to C$ $vec C$ with components

$C_{x} = A_{y} B_{z} - A_{z} B_{y}$ $C_x = A_yB_z - A_zB_y$

$C_{y} = A_{x} B_{z} - A_{z} B_{x}$ $C_y = A_xB_z - A_zB_x$

$C_{z} = A_{x} B_{y} - A_{y} B_{x}$ $C_z = A_xB_y - A_yB_x$

We have our components for vectors $\to A$ $vec A$ and $\to B$ $vec B$ expressed in their position vectors, and we'll use these values to calculate the components of $\to C$ $vec C$ :

$C_{x} = (5) (- 2) - (0) (1) = - 10$ $C_x = (5)(-2) - (0)(1) = -10$

$C_{y} = (4) (- 2) - (0) (4) = - 8$ $C_y = (4)(-2) - (0)(4) = -8$

$C_{z} = (4) (1) - (5) (4) = - 16$ $C_z = (4)(1) - (5)(4) = -16$

Using unit vectors, this vector product $\to C$ $vec C$ is

$(- 10) \to i + (- 8) \to j + (- 16) \to k$ $(-10)vec i + (-8)vec j + (-16)vec k$

Or, expressed as a position vector,

$< - 10, - 8, - 16 >$ $< -10, -8, -16>$

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What is the cross product of $< 4, 5, 0 >$ $<4 , 5 ,0 >$ and $< 4, 1, - 2 >$ $<4, 1 ,-2 >$ ?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

What is the cross product of <4,5,0><4 , 5 ,0 > and <4,1,−2><4, 1 ,-2 >?

1 Answer

Explanation:

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What is the cross product of $< 4, 5, 0 >$ $<4 , 5 ,0 >$ and $< 4, 1, - 2 >$ $<4, 1 ,-2 >$ ?