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# Find the equation of the plane passing through the point (−1,3,2) and perpendicular to each of the planes x+2y+3z=5 and 3x+3y+z=0.

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Solution

## The equation of plane passing through (x1,y1,z1) is given byA(x−x1)+B(y−y1)+C(z−z1)=0where, A,B,C are the direction ratios of normal to the plane.Now the plane passes through (−1,3,2)So, equation of plane is A(x+1)+B(y−3)+C(z−2)=0.......(1)Also, the plane is perpendicular to the given two planes.So, their normal to plane would be perpendicular to normals of both planes.We know that→a+→b is perpendicular to both →a and →bSo, required normal is cross product of normals of plane x+2y+3z=5 and 3x+3y+z=0Required normal =∣∣ ∣ ∣∣^i^j^k123331∣∣ ∣ ∣∣ =^i(2×1−3×3)−^j(1×1−3×3)+^k(1×3−3×2) =^i(2−9)−^j(1−9)+^k(3−6) =−7^i+8^j−3^kHence, direction ratios =(−7,8,−3)∴A=−7, B=8, C=−3Putting above values in (1)A(x+1)+B(y−3)+C(z−2)=0⇒−7(x+1)+8(y−3)−3(z−2)=0⇒−7x−7+8y−24−3z+6=0⇒−7x+8y−3z−25=0⇒7x−8y+3z+25=0Therefore equation of the required plane is 7x−8y+3z+25=0  Suggest Corrections  0      Similar questions  Related Videos   Introduction to Statistics
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