Question

Let \( \vec{a}(t)=\langle 4,-3,1\rangle t+\langle-6,5,6\rangle \) and \( \vec{b}(u)=\langle 0,3,-5\rangle u+\langle 6,-3,-9\rangle \). Find the acute angle (in degrees) between the lines: \( \square \)

          Let \( \vec{a}(t)=\langle 4,-3,1\rangle t+\langle-6,5,6\rangle \) and \( \vec{b}(u)=\langle 0,3,-5\rangle u+\langle 6,-3,-9\rangle \).

Find the acute angle (in degrees) between the lines: \( \square \)

Let a⃗(t)=⟨ 4,-3,1⟩ t+⟨-6,5,6⟩ and b⃗(u)=⟨ 0,3,-5⟩ u+⟨ 6,-3,-9⟩.

Find the acute angle (in degrees) between the lines: □

Added by Deborah K.

Calculus and Its Applications

Larry J. Goldstein, David C. Lay, David… 14th Edition

Chapter 8

Instant Answer

Solved by Expert Eleni Katirtzoglou

Step 1

The direction vector of \( \vec{a}(t) \) is \( \langle 4, -3, 1 \rangle \) and the direction vector of \( \vec{b}(u) \) is \( \langle 0, 3, -5 \rangle \). Show more…

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Let \( \vec{a}(t)=\langle 4,-3,1\rangle t+\langle-6,5,6\rangle \) and \( \vec{b}(u)=\langle 0,3,-5\rangle u+\langle 6,-3,-9\rangle \). Find the acute angle (in degrees) between the lines: \( \square \)