Suggested languages for you:

Americas

Europe

Q13E

Expert-verified
Found in: Page 572

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# (a) Find the symmetric equations for the line that passes through the point $$(1, - 5,6)$$ and parallel to the vector $$\langle - 1,2, - 3\rangle$$. (b) Find the point at which the line (that passes through the point $$(1, - 5,6)$$ and parallel to the vector $$\langle - 1,2, - 3\rangle )$$ intersects the $$xy$$-plane, the point at which the line (that passes through the point $$(1, - 5,6)$$ and parallel to the vector $$\langle - 1,2, - 3\rangle )$$ intersects the $$yz$$-plane and the point at which the line (that passes through the point $$(1, - 5,6)$$ and parallel to the vector $$\langle - 1,2, - 3\rangle )$$ intersects the $$xz$$-plane.

(a) The symmetric equations for the line that passes through the point and parallel to the vector $$\langle - 1,2, - 3\rangle$$ are $$\frac{{x - 1}}{{ - 1}} = \frac{{y + 5}}{2} = \frac{{z - 6}}{{ - 3}}$$.

(b) The point at which the line intersects the $$xy$$-plane is $$( - 1, - 1,0)$$.

The point at which the line intersects the $$yz$$ -plane is $$(0, - 3,3)$$.

The point at which the line intersects the $$xz$$-plane is $$\left( { - \frac{3}{2},0, - \frac{3}{2}} \right)$$.

See the step by step solution

## Step 1: Write the expression to find direction vector.

(a)

Formula used

$$\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}(1)$$

## Step 2: Calculate the symmetric equation.

In equation (1), substitute 1 for $${x_0}, - 5$$ for $${y_0},6$$ for $${z_0}, - 1$$ for $$a$$, $$2$$ for $$b$$, and $$- 3$$ for $$c$$.

$$\begin{array}{l}\frac{{x - 1}}{{ - 1}} = \frac{{y - ( - 5)}}{2} = \frac{{z - 6}}{{ - 3}}\frac{{x - 1}}{{ - 1}}\\\frac{{x - 1}}{{ - 1}} = \frac{{y + 5}}{2} = \frac{{z - 6}}{{ - 3}}\end{array}$$

As a result, the symmetric equations for the line that passes through the point $$(1, - 5,6)$$ and parallel to the vector $$\langle - 1,2, - 3\rangle$$ are $$\underline {\frac{{x - 1}}{{ - 1}} = \frac{{y + 5}}{2} = \frac{{z - 6}}{{ - 3}}}$$.

## Step 3: Formula and calculation of -coordinate.

(b)

Write the expression to find $$x$$-coordinate of point from the parametric equations as follows.

$$\frac{{x - 1}}{{ - 1}} = \frac{{z - 6}}{{ - 3}}$$

Rewrite the expression as.

$$x = - \left( {\frac{{z - 6}}{{ - 3}}} \right) + 1$$

In the expression, substitute 0 for $$z$$.

$$\begin{array}{l}x = - \left( {\frac{{0 - 6}}{{ - 3}}} \right) + 1\\x = - 2 + 1\\x = - 1\end{array}$$

## Step 4: Formula and calculation of $$y$$-coordinate.

Write the expression to find $$y$$-coordinate of point from the parametric equations as follows.

$$\frac{{y + 5}}{2} = \frac{{z - 6}}{{ - 3}}$$

Rewrite the expression as.

$$y = 2\left( {\frac{{z - 6}}{{ - 3}}} \right) - 5$$

In the expression, substitute 0 for $$z$$.

$$\begin{array}{l}y = 2\left( {\frac{{0 - 6}}{{ - 3}}} \right) - 5\\y = 4 - 5\\y = - 1\end{array}$$

The point at which the line intersects the $$xy$$- plane is $$( - 1, - 1,0)$$.

The line intersects the $$yz$$- plane when the value of $$x$$ $$(x = 0)$$

Write the expression to find -coordinate of point from the parametric equations as follows.

$$\frac{{x - 1}}{{ - 1}} = \frac{{y + 5}}{2}$$

Rewrite the expression as follows.

$$y = 2\left( {\frac{{x - 1}}{{ - 1}}} \right) - 5$$

In the expression, substitute 0 for $$x$$.

$$\begin{array}{l}y = 2\left( {\frac{{0 - 1}}{{ - 1}}} \right) - 5\\y = 2 - 5\\y = - 3\end{array}$$

## Step 5: Formula and calculation of $$z$$-coordinate.

Write the expression to find $$z$$-coordinate of point from the parametric equations as follows.

$$\frac{{x - 1}}{{ - 1}} = \frac{{z - 6}}{{ - 3}}$$

Rewrite the expression as follows.

$$z = ( - 3)\left( {\frac{{x - 1}}{{ - 1}}} \right) + 6$$

In the expression, substitute 0 for $$x$$.

$$\begin{array}{l}z = ( - 3)\left( {\frac{{0 - 1}}{{ - 1}}} \right) + 6\\z = - 3 + 6\\z = 3\end{array}$$

The point at which the line intersects the $$yz$$- plane is $$0, - 3,0)$$.

The line intersects the $$xz$$- plane when the value of $$x$$ $$(y = 0)$$

## Step 6: Formula and calculation of $$x$$-coordinate.

Write the expression to find $$x$$-coordinate of point from the parametric equations as follows.

$$\frac{{x - 1}}{{ - 1}} = \frac{{y + 5}}{2}$$

Rewrite the expression as follows.

$$x = ( - 1)\left( {\frac{{y + 5}}{2}} \right) + 1$$

In the expression, substitute 0 for $$y$$.

$$\begin{array}{l}x = ( - 1)\left( {\frac{{0 + 5}}{2}} \right) + 1\\x = - \frac{5}{2} + 1\\x = - \frac{3}{2}\end{array}$$

## Step 7: Formula and calculation of $$x$$-coordinate.

Write the expression to find $$z$$-coordinate of point from the parametric equations as follows.

$$\frac{{y + 5}}{2} = \frac{{z - 6}}{{ - 3}}$$

Rewrite the expression as follows.

$$z = ( - 3)\left( {\frac{{y + 5}}{2}} \right) + 6$$

In the expression, substitute 0 for $$y$$.

$$\begin{array}{l}z = ( - 3)\left( {\frac{{0 + 5}}{2}} \right) + 6\\z = - \frac{{15}}{2} + 6\\z = - \frac{3}{2}\end{array}$$

As a result, the point at which the line intersects the $$xz$$-plane is$$\left( { - \frac{3}{2},0, - \frac{3}{2}} \right)$$.