Question

Find the distance from the point Q(5, -2, 3) to the plane 4x - y + 4z = 8. (Note that if P is a point on the plane ax + by + cz = d, then the distance from any point Q to the plane equals the length of the orthogonal projection of $\overrightarrow{PQ}$ onto a vector $\mathbf{n} = (a, b, c)$ normal to the plane. This can be expressed as $d = \frac{|\overrightarrow{PQ} \cdot \mathbf{n}|}{|\mathbf{n}|}$.) The distance is

          Find the distance from the point Q(5, -2, 3) to the plane 4x - y + 4z = 8.
(Note that if P is a point on the plane ax + by + cz = d, then the distance from any point Q to the plane equals the length of the orthogonal projection of $\overrightarrow{PQ}$ onto a vector $\mathbf{n} = (a, b, c)$ normal to the plane. This can be expressed as $d = \frac{|\overrightarrow{PQ} \cdot \mathbf{n}|}{|\mathbf{n}|}$.)
The distance is

Find the distance from the point Q(5, -2, 3) to the plane 4x - y + 4z = 8.
(Note that if P is a point on the plane ax + by + cz = d, then the distance from any point Q to the plane equals the length of the orthogonal projection of PQ onto a vector 𝐧 = (a, b, c) normal to the plane. This can be expressed as d = (|PQ·𝐧|)/(|𝐧|).)
The distance is

Added by Sandra C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/27/2023

Step 1

The normal vector to the plane is n = (4, -1, 4). Show more…

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Find the distance from the point Q(5, -2, 3) to the plane 4x - y + 4z = 8. (Note that if P is a point on the plane ax + by + cz = d, then the distance from any point Q to the plane equals the length of the orthogonal projection of PQ onto a vector n = (a,b,c) normal to the plane. This can be expressed as d = $$\frac{|PQ \cdot n|}{|n|}$$). The distance is