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What is the distance of the point (1, 2, 3) form the plane x – 3y + 2z + 13 = 0?
- a)√14
- b)√13
- c)√11
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What is the distance of the point (1, 2, 3) form the plane x – 3...
Distance of a point from a plane:
The distance between a point and a plane can be calculated using the formula:
\[ \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Where (x1, y1, z1) is the coordinates of the point and Ax + By + Cz + D = 0 is the equation of the plane.
Given information:
Point P(1, 2, 3)
Equation of plane: x - 3y + 2z + 13 = 0
A = 1, B = -3, C = 2, D = 13
Calculating the distance:
Substitute the values into the formula:
\[ \frac{|1(1) + (-3)(2) + 2(3) + 13|}{\sqrt{1^2 + (-3)^2 + 2^2}} \]
\[ = \frac{|1 - 6 + 6 + 13|}{\sqrt{1 + 9 + 4}} \]
\[ = \frac{|14|}{\sqrt{14}} \]
\[ = \frac{14}{\sqrt{14}} \]
\[ = \sqrt{14} \]
Therefore, the distance of the point (1, 2, 3) from the plane x - 3y + 2z + 13 = 0 is \( \sqrt{14} \). Hence, option 'A' is the correct answer.
The distance between a point and a plane can be calculated using the formula:
\[ \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Where (x1, y1, z1) is the coordinates of the point and Ax + By + Cz + D = 0 is the equation of the plane.
Given information:
Point P(1, 2, 3)
Equation of plane: x - 3y + 2z + 13 = 0
A = 1, B = -3, C = 2, D = 13
Calculating the distance:
Substitute the values into the formula:
\[ \frac{|1(1) + (-3)(2) + 2(3) + 13|}{\sqrt{1^2 + (-3)^2 + 2^2}} \]
\[ = \frac{|1 - 6 + 6 + 13|}{\sqrt{1 + 9 + 4}} \]
\[ = \frac{|14|}{\sqrt{14}} \]
\[ = \frac{14}{\sqrt{14}} \]
\[ = \sqrt{14} \]
Therefore, the distance of the point (1, 2, 3) from the plane x - 3y + 2z + 13 = 0 is \( \sqrt{14} \). Hence, option 'A' is the correct answer.
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Community Answer
What is the distance of the point (1, 2, 3) form the plane x – 3...
Concept:
Perpendicular Distance of a Point from a Plane
Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d
And a point whose coordinate is, (x1, y1, z1)
Now, distance = 
Calculation:
Calculation:
We have to find the distance of the point (1, 2, 3) form the plane x – 3y + 2z + 13 = 0
Distance = 
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