Answer to Question #200628 in Analytic Geometry for Vina

Question #200628

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. 


1
Expert's answer
2021-06-08T06:51:37-0400

(5.1)


dAB=(32)2+(56)2+(7(1)2d_{AB}=\sqrt{(-3-2)^2+(-5-6)^2+(7-(-1)^2}

=210 (units)=\sqrt{210} \ (units)

(5.2)


2u3v+112w\vec 2u-3\vec v+1{1 \over 2}\vec w

=4,46+3,9,12+3,9,3=\langle-4,4-6\rangle+\langle3,-9,12\rangle+\langle3,-9,3\rangle

=2,14,9=\langle2,-14,9\rangle

2u3v+112w\|\vec 2u-3\vec v+1{1 \over 2}\vec w\|

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

(5.3)


w=(2)2+(6)2+(2)2=211\|\vec w\|=\sqrt{(2)^2+(-6)^2+(2)^2}=2\sqrt{11}

ww=12113,9,3\dfrac{\vec w}{\|\vec w\|}=\dfrac{1}{2\sqrt{11}}\langle3,-9,3\rangle

=31122,91122,31122=\langle\dfrac{3\sqrt{11}}{22},-\dfrac{9\sqrt{11}}{22},\dfrac{3\sqrt{11}}{22}\rangle

(5.4)


(u×v)w=u1u2u3v1v2v3w1w2w3(\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}

=u1v2v3w1w3u2v1v3w1w3+u3v1v2w1w2=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}



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