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Q.Find the equation of the plane passing through the intersection of the planes 2x+3y−z+1=02x + 3y - z + 1 = 0 and x+y−2z+3=0x + y - 2z + 3 = 0 and perpendicular to the plane 3x−y−2z−4=03x - y - 2z - 4 = 0. OR Find the image of the point (1,2,3)(1, 2, 3) in the plane x+2y+4z=38x + 2y + 4z = 38.

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2018Subjective· 4mImportance★★★★★
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Write the family of planes through the given line of intersection with a parameter kk, then fix kk by requiring the normal to be perpendicular to the third plane's normal.

The plane through the intersection of 2x+3y−z+1=02x+3y-z+1=0 and x+y−2z+3=0x+y-2z+3=0 can be written as

(2x+3y−z+1)+k(x+y−2z+3)=0(2x+3y-z+1)+k(x+y-2z+3)=0

(2+k)x+(3+k)y+(−1−2k)z+(1+3k)=0(∗)(2+k)x+(3+k)y+(-1-2k)z+(1+3k)=0 \qquad (\ast)

Its normal is n⃗1=(2+k, 3+k, −1−2k)\vec n_1=(2+k,\ 3+k,\ -1-2k). The given plane 3x−y−2z−4=03x-y-2z-4=0 has normal n⃗2=(3,−1,−2)\vec n_2=(3,-1,-2).

For (∗)(\ast) to be perpendicular to this plane, n⃗1⋅n⃗2=0\vec n_1\cdot\vec n_2=0:

3(2+k)+(−1)(3+k)+(−2)(−1−2k)=03(2+k)+(-1)(3+k)+(-2)(-1-2k)=0

(6+3k)+(−3−k)+(2+4k)=0(6+3k)+(-3-k)+(2+4k)=0

5+6k=0 ⇒ k=−565+6k=0\ \Rightarrow\ k=-\dfrac56

Substitute back into (∗)(\ast):

2+k=2−56=76,3+k=3−56=136,−1−2k=−1+53=23,1+3k=1−52=−322+k=2-\dfrac56=\dfrac76,\quad 3+k=3-\dfrac56=\dfrac{13}6,\quad -1-2k=-1+\dfrac53=\dfrac23,\quad 1+3k=1-\dfrac52=-\dfrac32

76x+136y+23z−32=0\dfrac76 x+\dfrac{13}6 y+\dfrac23 z-\dfrac32=0

Multiply through by 66:

7x+13y+4z−9=07x+13y+4z-9=0

Check: normal (7,13,4)⋅(3,−1,−2)=21−13−8=0(7,13,4)\cdot(3,-1,-2)=21-13-8=0 ✓ — perpendicular to 3x−y−2z−4=03x-y-2z-4=0, and it belongs to the family so it passes through the line of intersection of the first two planes.

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