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

(5.1) The distance between the two points.

(5.2) ||2~u − 3~v + (1) 1 2w~ ||.

(5.3) The unit vector in the direction of w~ .

(5.4) Suppose ~u; ~v and w~ are vectors in 3D, where ~u = (u1, u2, u3) ; ~v = (v1, v2, v3) and w~ =(w1, w2, w3).

Express (~u × ~v) · w~ as a determinant. 

Answer 1

(5.1)

$$d_{AB}=\sqrt{(-3-2)^2+(-5-6)^2+(7-(-1)^2}$$

$$=\sqrt{210} \ (units)$$

(5.2)

$$\vec 2u-3\vec v+1{1 \over 2}\vec w$$

$$=\langle-4,4-6\rangle+\langle3,-9,12\rangle+\langle3,-9,3\rangle$$

$$=\langle2,-14,9\rangle$$

$$\|\vec 2u-3\vec v+1{1 \over 2}\vec w\|$$

$$=\sqrt{(2)^2+(-14)^2+(9)^2}=\sqrt{281}$$

(5.3)

$$\|\vec w\|=\sqrt{(2)^2+(-6)^2+(2)^2}=2\sqrt{11}$$

$$\dfrac{\vec w}{\|\vec w\|}=\dfrac{1}{2\sqrt{11}}\langle3,-9,3\rangle$$

$$=\langle\dfrac{3\sqrt{11}}{22},-\dfrac{9\sqrt{11}}{22},\dfrac{3\sqrt{11}}{22}\rangle$$

(5.4)

$$(\vec u\times \vec v)\cdot\vec w= \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \\ \end{vmatrix}$$

$$=u_1\begin{vmatrix} v_2 & v_3 \\ w_1 & w_3 \end{vmatrix}-u_2\begin{vmatrix} v_1 & v_3 \\ w_1 & w_3 \end{vmatrix}+u_3\begin{vmatrix} v_1 & v_2 \\ w_1 & w_2 \end{vmatrix}$$