Question

A vector field is specified as \(\mathbf{G}=24 x y \mathbf{a}_{x}+12\left(x^{2}+2\right) \mathbf{a}_{y}+18 z^{2} \mathbf{a}_{z} .\) Given two points, \(P(1,2,-1)\) and \(Q(-2,1,3)\), find \((a) \mathbf{G}\) at \(P ;(b)\) a unit vector in the direction of \(\mathbf{G}\) at \(Q ;(c)\) a unit vector directed from \(Q\) toward \(P ;(d)\) the equation of the surface on which \(|\mathbf{G}|=60\).

Short answer

(b) Find a unit vector in the direction of the vector field \(\mathbf{G}\) at point \(Q (-2, 1, 3)\). (c) What is a unit vector directed from point \(Q\) towards point \(P\)? (d) Determine the equation of the surface on which the magnitude of the vector field \(\mathbf{G}\) is equal to 60.

Step-by-step solution

(a) Find \(\mathbf{G}\) at point \(P\)(1,2,-1)

To find the value of \(\mathbf{G}\) at point \(P\), substitute the coordinates of point \(P\) into the expressions for \(\mathbf{G}\): \(\mathbf{G}(x,y,z) = 24xy\,\mathbf{a}_x + 12(x^2+2)\,\mathbf{a}_y + 18z^2\,\mathbf{a}_z\). So, \(\mathbf{G}(1,2,-1) = 24(1)(2)\,\mathbf{a}_x + 12((1)^2+2)\,\mathbf{a}_y + 18(-1)^2\,\mathbf{a}_z = 48\,\mathbf{a}_x + 36\,\mathbf{a}_y + 18\,\mathbf{a}_z\).

(b) Find a unit vector in the direction of \(\mathbf{G}\) at point \(Q\)(-2,1,3)

First, we need to find the value of \(\mathbf{G}\) at point \(Q\). Substitute the coordinates of point \(Q\) into the expressions for \(\mathbf{G}\): \(\mathbf{G}(-2,1,3) = 24(-2)(1)\,\mathbf{a}_x + 12((-2)^2+2)\,\mathbf{a}_y + 18(3)^2\,\mathbf{a}_z = -48\,\mathbf{a}_x + 60\,\mathbf{a}_y + 162\,\mathbf{a}_z\). Now we will find the magnitude of \(\mathbf{G}\) at point \(Q\): \(|\mathbf{G}| = \sqrt{(-48)^2 + 60^2 + 162^2} = 6\sqrt{170}\). A unit vector in the direction of \(\mathbf{G}\) at point \(Q\): \(\mathbf{\hat{G}}_Q = \frac{\mathbf{G}}{|\mathbf{G}|} = \frac{1}{6\sqrt{170}}(-48\,\mathbf{a}_x + 60\,\mathbf{a}_y + 162\,\mathbf{a}_z)\).

(c) Find a unit vector directed from \(Q\) towards \(P\)

First, we need to find the vector from point \(Q\) to point \(P\): \(\mathbf{V}_{QP} = (1 - (-2))\,\mathbf{a}_x + (2 - 1)\,\mathbf{a}_y + (-1 - 3)\,\mathbf{a}_z = 3\,\mathbf{a}_x + \mathbf{a}_y - 4\,\mathbf{a}_z\). Now we will find the magnitude of this vector: \(|\mathbf{V}_{QP}| = \sqrt{(3)^2 + (1)^2 + (-4)^2} = \sqrt{26}\). A unit vector directed from \(Q\) towards \(P\): \(\mathbf{\hat{V}}_{QP} = \frac{\mathbf{V}_{QP}}{|\mathbf{V}_{QP}|} = \frac{1}{\sqrt{26}}(3\,\mathbf{a}_x + \mathbf{a}_y - 4\,\mathbf{a}_z)\).

(d) Find the equation of the surface on which \(|\mathbf{G}|=60\)

First, we find the magnitude of \(\mathbf{G}\) using the given expression for \(\mathbf{G}\): \(|\mathbf{G}| = \sqrt{(24xy)^2 + (12(x^2+2))^2 + (18z^2)^2}\). Now we will equate the magnitude of \(\mathbf{G}\) to 60: \(60 = \sqrt{(24xy)^2 + (12(x^2+2))^2 + (18z^2)^2}\). Square the equation to get rid of the square root: \(3600 = (24xy)^2 + (12(x^2+2))^2 + (18z^2)^2\). This is the equation of the surface on which the magnitude of \(\mathbf{G}\) is equal to 60.