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A straight line passes through (1, - 2 , 3 ) and perpendicular to the plane 2x + 3y - z = 7.​
What are the direction ratios of normal to plane ?
  • a)
    < 2 , 3 , - l >
  • b)
    < 2 , 3 , 1>
  • c)
    < - 1 , 2 , 3>
  • d)
    None o f these
Correct answer is option 'A'. Can you explain this answer?
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A straight line passes through (1, - 2 , 3 ) and perpendicular to the ...
Direction ratios o f normal to plane 2x + 3 y - z = 7 is < 2 ,3 ,- 1 >
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A straight line passes through (1, - 2 , 3 ) and perpendicular to the ...
Explanation:

Finding the normal vector to the plane:
Given plane equation: 2x + 3y - z = 7
The normal vector to the plane is the coefficients of x, y, and z in the plane equation. So, the normal vector is <2, 3,="" -1="">.

Understanding perpendicularity:
For a line to be perpendicular to a plane, the direction ratios of the line should be parallel to the normal vector of the plane. If two vectors are perpendicular, their dot product is zero.

Given point on the line:
The line passes through the point (1, -2, 3), which can be considered as a point on the line.

Direction ratios of the line:
To find the direction ratios of the line passing through the given point and perpendicular to the plane, we use the point on the line and the normal vector of the plane. The direction ratios will be the same as the normal vector of the plane, but with opposite signs.
So, the direction ratios of the line are <-2, -3,="" 1="">, which is the same as the normal vector of the plane but with opposite signs.
Therefore, the correct option is < 2 , 3 , -1 >.
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A straight line passes through (1, - 2 , 3 ) and perpendicular to the plane 2x + 3y - z = 7.​What are the direction ratios of normal to plane ?a)< 2 , 3 , - l >b)< 2 , 3 , 1>c)< - 1 , 2 , 3>d)None o f theseCorrect answer is option 'A'. Can you explain this answer?
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