Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![**Problem 19:** Find the distance from the point \((3, -2, 7)\) to the plane \(4x - 6y + z = 5\). Show all necessary work to receive full credit.
**Solution:**
To find the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\), use the formula:
\[
d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
**Given:**
- Point \((x_1, y_1, z_1) = (3, -2, 7)\)
- Plane equation \(4x - 6y + z = 5\)
First, rewrite the plane equation in the form \(Ax + By + Cz + D = 0\):
\[4x - 6y + z - 5 = 0\]
Thus,
- \(A = 4\)
- \(B = -6\)
- \(C = 1\)
- \(D = -5\)
Substitute these values into the distance formula:
\[
d = \frac{|4(3) - 6(-2) + 1(7) - 5|}{\sqrt{4^2 + (-6)^2 + 1^2}}
\]
Calculate the numerator:
\[
|4(3) - 6(-2) + 1(7) - 5| = |12 + 12 + 7 - 5| = |26|
\]
Calculate the denominator:
\[
\sqrt{4^2 + (-6)^2 + 1^2} = \sqrt{16 + 36 + 1} = \sqrt{53}
\]
Thus, the distance is:
\[
d = \frac{26}{\sqrt{53}}
\]
Therefore, the distance from the point \((3, -2, 7)\) to the plane \(4x - 6y + z = 5\) is \(\frac{26}{\sqrt{53}}\).](https://content.bartleby.com/qna-images/question/3ba91fa8-c4da-44a3-a037-923d1d08ef86/9f97895a-eeb8-4f66-956e-be45cbb0681e/b4iq8af_processed.png)
Transcribed Image Text:**Problem 19:** Find the distance from the point \((3, -2, 7)\) to the plane \(4x - 6y + z = 5\). Show all necessary work to receive full credit.
**Solution:**
To find the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\), use the formula:
\[
d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
**Given:**
- Point \((x_1, y_1, z_1) = (3, -2, 7)\)
- Plane equation \(4x - 6y + z = 5\)
First, rewrite the plane equation in the form \(Ax + By + Cz + D = 0\):
\[4x - 6y + z - 5 = 0\]
Thus,
- \(A = 4\)
- \(B = -6\)
- \(C = 1\)
- \(D = -5\)
Substitute these values into the distance formula:
\[
d = \frac{|4(3) - 6(-2) + 1(7) - 5|}{\sqrt{4^2 + (-6)^2 + 1^2}}
\]
Calculate the numerator:
\[
|4(3) - 6(-2) + 1(7) - 5| = |12 + 12 + 7 - 5| = |26|
\]
Calculate the denominator:
\[
\sqrt{4^2 + (-6)^2 + 1^2} = \sqrt{16 + 36 + 1} = \sqrt{53}
\]
Thus, the distance is:
\[
d = \frac{26}{\sqrt{53}}
\]
Therefore, the distance from the point \((3, -2, 7)\) to the plane \(4x - 6y + z = 5\) is \(\frac{26}{\sqrt{53}}\).
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