## University Calculus: Early Transcendentals (3rd Edition)

a) $-7j$; b) $-\dfrac{3 \sqrt 2}{5} i -\dfrac{4 \sqrt 2}{5}k$; c) $\dfrac{1}{4}i-\dfrac{1}{3}j-k$ and d) $\dfrac{a}{\sqrt 2}i+\dfrac{a}{\sqrt 3}j -\dfrac{a}{\sqrt 6}k$
Here, a) $7(-j)=-7j$; b) $\sqrt 2(-\dfrac{3}{5}i-\dfrac{4}{5}k)=-\dfrac{3 \sqrt 2}{5} i -\dfrac{4 \sqrt 2}{5}k$; c) $\dfrac{13}{12}(\dfrac{3}{13}i-\dfrac{4}{13}j-\dfrac{12}{13}k)=\dfrac{1}{4}i-\dfrac{1}{3}j-k$ and d) $a(\dfrac{1}{\sqrt 2}i+\dfrac{1}{\sqrt 3}j -\dfrac{1}{\sqrt 6}k)=\dfrac{a}{\sqrt 2}i+\dfrac{a}{\sqrt 3}j -\dfrac{a}{\sqrt 6}k$