Question

MATH2412.002 20FA MW 9-10:50 F 9-9:50 BB218 Homework: 8.6 Vectors in Space Score: 0 of 1 pt 8.6.59 8 of 12 (6 Find the direction angles of the given vector. Write the vector in terms of its magnitude and direction cosines as $v = \|v\| [ (\cos \alpha)i + (\cos \beta)j + (\cos \gamma)k]$. v = -12i - 3j + 4k $\alpha = $\boxed{} degrees (Round to the nearest tenth of a degree, if necessary.)

          MATH2412.002 20FA MW 9-10:50 F 9-9:50 BB218
Homework: 8.6 Vectors in Space
Score: 0 of 1 pt
8.6.59
8 of 12 (6
Find the direction angles of the given vector. Write the vector in terms of its
magnitude and direction cosines as $v = \|v\| [ (\cos \alpha)i + (\cos \beta)j + (\cos \gamma)k]$.
v = -12i - 3j + 4k
$\alpha = $\boxed{} degrees
(Round to the nearest tenth of a degree, if necessary.)

MATH2412.002 20FA MW 9-10:50 F 9-9:50 BB218
Homework: 8.6 Vectors in Space
Score: 0 of 1 pt
8.6.59
8 of 12 (6
Find the direction angles of the given vector. Write the vector in terms of its
magnitude and direction cosines as v = v [ (cosα)i + (cosβ)j + (cosγ)k].
v = -12i - 3j + 4k
α = degrees
(Round to the nearest tenth of a degree, if necessary.)

Added by Karen F.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert William Semus

Step 1

We can use the formula for the magnitude of a vector in three-dimensional space: |v| = sqrt(v1^2 + v2^2 + v3^2) Substituting the given values, we get: |v| = sqrt(12^2 + (-3)^2 + 4^2) = sqrt(169) = 13 Show more…

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MATH2412.002 ZOFA MW 9-10.50 F 9-9.50 BB218 Homework: 8.6 Vectors in Space Score: 0 of 1 pt 8 of 12 8.6.59 Find the direction angles of the given vector. Write the vector in terms of its magnitude and direction cosines as v = ||v|| [ (cos Î±)i + (cos Î²)j + (cos Î³)k ] 12i - 3j + 4k Î¸ = degrees [(Round to the nearest tenth of a degree, if necessary)]