Answer to Question #223387 in Linear Algebra for sammuel

Question

Let u=(-1/2,-1,2/3) ,v=(4,-1,-3) and θ=1/2. Find

i) u.θv

ii)A non-zero vector orthogonal to both u and v

Answer 1

u=(-1/2,-1,2/3)

v=(4,-1,-3)

u.v = ?

θ = ?

(i)

$u = -1/2i -j + 2/3k\\ |u| =\sqrt{(-1/2)²+( -1)²+ (2/3)²) }\\=1.302\\$

$v = 4i-j -k\\ |v|= \sqrt{(4)²+( -1)²+ (-3)²)} =5.10$

dot product (u.v) = $(a_ib_i + a_jb_j + a_kb_k)$

$= ((-1/2×4)+ (-1×-1) +(2/3×-3))\\ = (-2+1-2) = -3$

$\theta=\cos^{-1} \dfrac{u.v}{|u||v|} =\cos^{-1}\dfrac{-3}{1.302× 5.10}\\$

$=\cos^{-1} -0.25 = 116.9°$

(ii)

$u = -1/2i -j + 2/3k\\ v = 4i-j -3k$

A vector $C \neq O$ that is orthogonal to both.

Since $C = (c_1, c_2, c_3)$ must be orthogonal to both u and v;

$\begin{aligned} u\cdot C &= 0 & \implies & &-1/2c_1 - c_2 + 2/3c_3 &= 0 \\ v\cdot C &= 0 & \implies & &4c_1 -c_2 - 3c_3 = 0 \end{aligned}$

From the first equation; $c_2= -1/2c_1 +2/3c_3$

$[ 4c_1 -c_2 - 3c_3 = 0 \quad \implies \quad 4c_1 -(-1/2c_1 + 2/3c_3) - 3c_3= 0 \quad \implies \\ 4c_1+1/2c_1-2/3c_3-3c_3=0\\ \implies 9/2\ c_1 = 11/3\ c_3$

$\therefore c_1 = \dfrac{22}{27}\ c_3$

Since $c_3$ is then arbitrary we choose $c_3 = 27$ and obtain

$[ c_2 = 7 \qquad c_1 = 22]$

Therefore, $C = (22,7,27)$