Answer to Question #262384 in Analytic Geometry for Rick

Question

Let u = 2i – 3j + 4k v = 3i + k – 2k w = i+ 5j+3k

a. 2u + 3v + 4w

b. u . v and u . w

c. u x v and || u x v ||

d. Find the unit vector of u, v and w

e. Proj (u,w)

f. Establish the angle between the u and v.

Answer 1

(a) $2u+3v+4w = 2(2i – 3j + 4k) + 3(3i + j – 2k) + 4(i + 5j + 3k) = 4i-6j+8k+9i+3j-6k+4i+20j+12k = 17i+17j+14k$

(b) $u•v = (2i – 3j + 4k)•(3i + j – 2k) = 6-3-8=-5$

$u•w = (2i – 3j + 4k)•(i + 5j + 3k) = 2-15+12=-1$

(c) $u×v = \begin{pmatrix} i & j & k \\ 2 & -3 & 4 \\ 3 & 1 & -2 \end{pmatrix} = i(6-4)-j(-4-12)+k(2+9) = 2i+16j+11k$

$|u×v| = \sqrt{2²+16²+11²} = \sqrt{381}$

(d) unit vector of u$= \frac{u}{|u|} = \frac{2i – 3j + 4k}{\sqrt{2²+(-3)²+4²}} = \frac{2i – 3j + 4k}{\sqrt{29}}$

unit vector of v = $\frac{v}{|v|} = \frac{3i + j – 2k}{\sqrt{3²+1²+(-2)²}} = \frac{3i + j – 2k}{\sqrt{14}}$

unit vector of w = $\frac{w}{|w|} = \frac{i + 5j + 3k}{\sqrt{1²+5²+3²}} = \frac{i + 5j + 3k}{\sqrt{35}}$

(e) Proj(u,w) = $\frac{u•w}{|w|²} w = \frac{-1}{35} (i + 5j + 3k)$

(f) Angle between u and v

$Cos∅ = \frac{u•v}{|u||v|} = \frac{-5}{\sqrt{29×14}} = \frac{-5}{\sqrt{406}}$

$=> ∅ = cos^{-1}(\frac{-5}{\sqrt{406}})$

$∅ = 104.37°$