What is the unit vector that is orthogonal to the plane containing # (-2i- 3j + 2k) # and # (3i – 4j + 4k) #?

Question

What is the unit vector that is orthogonal to the plane containing #  (-2i- 3j + 2k) # and #  (3i – 4j + 4k)        #?

Julia Aguilar · Accepted Answer

Take the cross product of the 2 vectors
#v_1 = (-2, -3, 2) and v_2 = (3, -4, 4) #
Compute #v_3 = v_1 xx v_2 #
#1/sqrt(501) (-4, 14, 17)#

The #v_3 = (-4, 14, 17)#
The magnitude of this new vector is:
#|v_3| = 4^2 + 14^2 + 17^2#
Now to find the unit vector normalize our new vector
#u_3 = v_3/ (sqrt( 4^2 + 14^2 + 17^2)); = 1/sqrt(501) (-4, 14, 17)#

Elena Albarran · Answer

undefined To find the unit vector orthogonal to the plane containing the given vectors, first find the cross product of the two vectors. Then, normalize the resulting vector to obtain the unit vector.

Cross product: 
$$ \vec{v}_1 = (-2i- 3j + 2k) $$
$$ \vec{v}_2 = (3i - 4j + 4k) $$
$$ \vec{n} = \vec{v}_1 \times \vec{v}_2 $$

Normalize the resulting vector:
$$ |\vec{n}| = \sqrt{a^2 + b^2 + c^2} $$
$$ \vec{u} = \frac{\vec{n}}{|\vec{n}|} $$

Substitute the values of $ \vec{v}_1 $ and $ \vec{v}_2 $ into the cross product formula:
$$ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & -3 & 2 \ 3 & -4 & 4 \end{vmatrix} $$

Solve for $ \vec{n} $, then normalize it to find the unit vector orthogonal to the given plane.