Question:
If \((3,4,-7)\) is the foot of the perpendicular drawn from the point \((-2,3,6)\) to the plane \(\pi\), then the sum of the intercepts made by the plane \(\pi\) on the \(x\)- and \(y\)-axes is
If \((3,4,-7)\) is the foot of the perpendicular drawn from the point \((-2,3,6)\) to the plane \(\pi\), then the sum of the intercepts made by the plane \(\pi\) on the \(x\)- and \(y\)-axes is
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If a point is the foot of perpendicular from another point to a plane, then the line joining the point and its foot is normal to the plane.
Updated On: Jun 26, 2026
- \(132\)
- \(142\)
- \(210\)
- \(175\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the normal vector of the plane.
The point from which perpendicular is drawn is \[ A(-2,3,6) \] The foot of perpendicular is \[ P(3,4,-7) \] Since \(AP\) is perpendicular to the plane \(\pi\), the direction vector of \(AP\) is normal to the plane.
So, \[ \overrightarrow{AP}=(3+2,4-3,-7-6) \] \[ \overrightarrow{AP}=(5,1,-13) \] Thus, the normal vector of the plane is \[ (5,1,-13) \]
Step 2: Write the equation of the plane.
The plane passes through \[ P(3,4,-7) \] Using point-normal form, \[ 5(x-3)+1(y-4)-13(z+7)=0 \] Expanding, \[ 5x-15+y-4-13z-91=0 \] \[ 5x+y-13z-110=0 \]
Step 3: Find the intercept on the \(x\)-axis.
For \(x\)-intercept, put \[ y=0,\quad z=0 \] Then, \[ 5x-110=0 \] \[ x=22 \] So, the \(x\)-intercept is \[ 22 \]
Step 4: Find the intercept on the \(y\)-axis.
For \(y\)-intercept, put \[ x=0,\quad z=0 \] Then, \[ y-110=0 \] \[ y=110 \] So, the \(y\)-intercept is \[ 110 \]
Step 5: Find the required sum.
The required sum is \[ 22+110=132 \]
Step 6: Final conclusion.
Therefore, the sum of intercepts made by the plane on the \(x\)- and \(y\)-axes is \[ \boxed{132} \]
The point from which perpendicular is drawn is \[ A(-2,3,6) \] The foot of perpendicular is \[ P(3,4,-7) \] Since \(AP\) is perpendicular to the plane \(\pi\), the direction vector of \(AP\) is normal to the plane.
So, \[ \overrightarrow{AP}=(3+2,4-3,-7-6) \] \[ \overrightarrow{AP}=(5,1,-13) \] Thus, the normal vector of the plane is \[ (5,1,-13) \]
Step 2: Write the equation of the plane.
The plane passes through \[ P(3,4,-7) \] Using point-normal form, \[ 5(x-3)+1(y-4)-13(z+7)=0 \] Expanding, \[ 5x-15+y-4-13z-91=0 \] \[ 5x+y-13z-110=0 \]
Step 3: Find the intercept on the \(x\)-axis.
For \(x\)-intercept, put \[ y=0,\quad z=0 \] Then, \[ 5x-110=0 \] \[ x=22 \] So, the \(x\)-intercept is \[ 22 \]
Step 4: Find the intercept on the \(y\)-axis.
For \(y\)-intercept, put \[ x=0,\quad z=0 \] Then, \[ y-110=0 \] \[ y=110 \] So, the \(y\)-intercept is \[ 110 \]
Step 5: Find the required sum.
The required sum is \[ 22+110=132 \]
Step 6: Final conclusion.
Therefore, the sum of intercepts made by the plane on the \(x\)- and \(y\)-axes is \[ \boxed{132} \]
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