Question

Consider the vectors u=<-2,2, - 3>, v=<-1,3, - 4>, w=2, - 6,2> and the points A (2,6,-1) and B(-3, - 5,7). Evaluate

a.) The distance between the two points

b.) ||2u - 3v + 1/2w||

c.) The unit vector in the direction of w.

d.) Suppose u; v and w are vectors in 3D,where u=(u_{1, }u₂, u₃) ; v = (v₁, v₂, v₃) and w=(w₁, w₂, w₃).

Express (u x v). W as a determinant.

Answer 1

Consider the vectors $u=(-2,2, - 3)$, $v=(-1,3, - 4), w=(2, - 6,2)$ and the points $A (2,6,-1)$ and $B(-3, - 5,7)$.

a) The distance between the two points is equal to $\sqrt{(-3-2)^2+(-5-6)^2+(7-(-1))^2}=\sqrt{25+121+64}=\sqrt{210}\approx14.49$

b.) $||2u - 3v + \frac{1}{2}w|| =||2(-2,2,-3)-3(-1,3,-4)+\frac{1}{2}(2,-6,2)||=||(0,-8,7)||=\sqrt{(-8)^2+7^2}=\sqrt{113}\approx 10.63$

с) The unit vector in the direction of $w$ is $\frac{w}{||w||}=\frac{(2,-6,2)}{\sqrt{4+36+4}}=\frac{(2,-6,2)}{\sqrt{44}}=\frac{2(1,-3,1)}{2\sqrt{11}}=(\frac{1}{\sqrt{11}},-\frac{3}{\sqrt{11}},\frac{1}{\sqrt{11}}).$

d.) Suppose $u, v$ and $w$ are vectors in 3D, where $u=(u_1, u_2, u_3),\ v = (v_1, v_2, v_3)$ and $w=(w_1, w_2, w_3)$.

$(u \times v)\cdot w=(u_2v_3-v_2u_3,v_1u_3-u_1v_3,u_1v_2-v_1u_2)\cdot(w_1,w_2,w_3)=(u_2v_3-v_2u_3)w_1+(v_1u_3-u_1v_3)w_2+(u_1v_2-v_1u_2)w_3=\begin{vmatrix}u_1 & u_ 2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{vmatrix}.$