Chapter 9, Problem 7T

(a)

To determine

The distance between the points $P\left(4,3,-1\right)$ and $Q\left(6,-1,3\right)$.

Answer to Problem 7T

The distance between the points $P\left(4,3,-1\right)$ and $Q\left(6,-1,3\right)$ is 6.

Explanation of Solution

Given:

The two points are $P\left(4,3,-1\right)$ and $Q\left(6,-1,3\right)$.

Formula used: The distance formula for the points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ is given below,

$d\left(P,Q\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$ (1)

Calculation:

Substitute 4 for $x_1$, 3 for $y_1$, $-1$ for $z_1$, 6 for $x_2$, $-1$ for $y_2$ and 3 for $z_2$ in equation (1), to find the distance between the points $P\left(4,3,-1\right)$ and $Q\left(6,-1,3\right)$,

$\begin{array}{c}d\left(P,Q\right)=\sqrt{{\left(6-4\right)}^2+{\left(-1-3\right)}^2+{\left(3-\left(-1\right)\right)}^2}\\ =\sqrt{2^2+{\left(-4\right)}^2+{\left(3+1\right)}^2}\\ =\sqrt{4+16+4^2}\end{array}$

Simplify the above expression,

$\begin{array}{c}d\left(P,Q\right)=\sqrt{20+16}\\ =\sqrt{36}\\ =6\end{array}$

Thus, the distance between the points $P\left(4,3,-1\right)$ and $Q\left(6,-1,3\right)$ is 6.

(b)

To determine

The equation of the sphere whose center is $P\left(4,3,-1\right)$ and the segment $\overrightarrow{PQ}$ is the radius of the sphere.

Answer to Problem 7T

The equation of the sphere whose center is $P\left(4,3,-1\right)$ and the segment $\overrightarrow{PQ}$ is the radius of the sphere is $x^2+y^2+z^2-8x-6y+2z=10$.

Explanation of Solution

Given:

The radius $r$ of the sphere is the segment $\overrightarrow{PQ}$.

The center of the sphere is $P\left(4,3,-1\right)$.

Formula used: The equation of the sphere with radius $r$ and center $C\left(h,k,l\right)$ is given below,

${\left(x-h\right)}^2+{\left(y-k\right)}^2+{\left(z-l\right)}^2=r^2$ (2)

Calculation:

From part (a), the value of segment $\overrightarrow{PQ}$ is 6, therefore the radius of sphere is 6.

The radius of the sphere is 6 and the center is $P\left(4,3,-1\right)$.

Substitute 6 for $r$, 4 for h, 3 for k and $-1$ for l in equation (2), to find the equation of the sphere,

$\begin{array}{c}{\left(x-4\right)}^2+{\left(y-3\right)}^2+{\left(z-\left(-1\right)\right)}^2={\left(6\right)}^2\\ \left(x^2-8x+16\right)+\left(y^2-6y+9\right)+\left(z^2+2z+1\right)=36\\ x^2+y^2+z^2-8x-6y+2z+26=36\\ x^2+y^2+z^2-8x-6y+2z=10\end{array}$

Thus, the equation of the sphere whose center is $P\left(4,3,-1\right)$ and the segment $\overrightarrow{PQ}$ is the radius of the sphere is $x^2+y^2+z^2-8x-6y+2z=10$.

(c)

To determine

The vector u with initial point $P\left(4,3,-1\right)$ and terminal point $Q\left(6,-1,3\right)$ in component form and in terms of i, j and k.

Answer to Problem 7T

The component form of u is $\langle 2,-4,4\rangle$ and the vector u in terms of i, j and k is $2i-4j+4k$.

Explanation of Solution

Given:

The initial point is $P\left(4,3,-1\right)$ and the terminal point is $Q\left(6,-1,3\right)$.

Formula used: The vector with initial point $A\left(a_1,a_2,a_3\right)$ and terminal point $B\left(b_1,b_2,b_3\right)$ is given below,

$\overrightarrow{AB}=\langle b_1-a_1,b_2-a_2,b_3-a_3\rangle$ (3)

The vector $v=\langle a_1,a_2,a_3\rangle$ in terms of i, j and k is given below,

$v=\langle a_1,a_2,a_3\rangle =a_{1}i+a_{2}j+a_{3}k$ (4)

Calculation:

Substitute 4 for $a_1$, 3 for $a_2$, $-1$ for $a_3$, 6 for $b_1$, $-1$ for $b_2$ and 3 for $b_3$ in equation (3), to find the vector u,

$\begin{array}{c}u=\langle 6-4,-1-3,3-\left(-1\right)\rangle \\ =\langle 2,-4,3+1\rangle \\ =\langle 2,-4,4\rangle \end{array}$

Substitute 2 for $a_1$, $-4$ for $a_2$ and 4 for $a_3$ in equation (4), to find the vector u in terms of i, j and k,

$u=\langle 2,-4,4\rangle =2i-4j+4k$

Thus, the component form of u is $\langle 2,-4,4\rangle$ and the vector u in terms of i, j and k is $2i-4j+4k$.