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Q13E

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Essential Calculus: Early Transcendentals
Found in: Page 572
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

(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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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\(\left( { - \frac{3}{2},0, - \frac{3}{2}} \right)\).

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