What is the angle between $< 6, 9, - 2 >$ $<6,9,-2>$ and $< - 2, 0, 4 >$ $< -2,0,4 >$ ?

Question

What is the angle between $< 6, 9, - 2 >$ $<6,9,-2>$ and $< - 2, 0, 4 >$ $< -2,0,4 >$ ?

1 Answer

Impact of this question

1537 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-12T17:20:05

$θ = arccos \frac{a x + b y + c z}{\sqrt{x^{2} + y^{2} + z^{2}} \sqrt{a^{2} + b^{2} + c^{2}}}$ $theta = arccos frac{ax + by + cz}{ sqrt {x^2 + y^2 + z^2} sqrt {a^2 + b^2 + c^2}}$

Explanation:

There are many multiplicities of this question, so let me solve all of them.

$v = (x, y, z)$ $v = (x, y, z)$

$w = (a, b, c)$ $w = (a, b, c)$

$v \cdot w = | v | \cdot | w | \cdot cos θ$ $v * w = |v| * |w| * cos theta$

$a x + b y + c z = \sqrt{x^{2} + y^{2} + z^{2}} \sqrt{a^{2} + b^{2} + c^{2}} cos θ$ $ax + by + cz = sqrt {x^2 + y^2 + z^2} sqrt {a^2 + b^2 + c^2} cos theta$

What is left is your arithmetical hability!

$- 12 - 8 = \sqrt{36 + 81 + 4} \sqrt{4 + 16} cos θ$ $-12 -8 = sqrt {36 + 81 + 4} sqrt {4 + 16} cos theta$

$- \frac{20}{11 \cdot \sqrt{20}} = cos θ = - \frac{\sqrt{20}}{11}$ $-20 / (11 * sqrt20) = cos theta = - sqrt 20 / 11$

Science

Math

Humanities

... and beyond

What is the angle between $< 6, 9, - 2 >$ $<6,9,-2>$ and $< - 2, 0, 4 >$ $< -2,0,4 >$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the angle between <6,9,−2><6,9,-2> and <−2,0,4>< -2,0,4 >?

1 Answer

Explanation:

Related questions

Impact of this question

What is the angle between $< 6, 9, - 2 >$ $<6,9,-2>$ and $< - 2, 0, 4 >$ $< -2,0,4 >$ ?