Answer to Question #118540 in Analytic Geometry for Joshua

Question

Given α=i+2j+3k, β=2i+j−k find α×β
a.5i+7j-3k
b.-5i+7j-3k
c.-5i+7j+3k
d.-5i-7j-3k

Answer 1

$\alpha = i + 2j + 3k, \beta = 2i+j-k$

$*$ indicates the cross product

$\alpha$ x $\beta$ = $(i+2j+3k)$ * $(2i+j-k)$

$= 2(i*i) + (i*j) - (i*k) + 4(j*i) + 2(j*j) -2(j*k) +6(k*i) + 3(k*j) - 3(k*k)$

since $i*i = j*j = k*k = 0, i*j = k, j*k = i, k*i = j, j*i = - k , k*j = -i,$

$i*k = -j$

so $\alpha$ x $\beta$ $= 2(0) + k + j -4k +2(0) - 2i + 6j -3i -3(0)$

$= -5i + 7j - 3k$

So option b is correct.