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Q.Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.

Goa GbshseGBSHSE Class 12 Board Exam 2018Subjective· 4mImportance★★★★★
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The required plane's normal is the cross product of the two given planes' normals; then use the point to fix the constant.

The two given planes have normals n⃗1=(2,3,−2)\vec n_1=(2,3,-2) and n⃗2=(1,2,−3)\vec n_2=(1,2,-3). A plane perpendicular to both must have a normal along n⃗1×n⃗2\vec n_1\times\vec n_2:

n⃗1×n⃗2=∣i^j^k^23−212−3∣=i^(3(−3)−(−2)(2))−j^(2(−3)−(−2)(1))+k^(2(2)−3(1))\vec n_1\times\vec n_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\2&3&-2\\1&2&-3\end{vmatrix} = \hat i\big(3(-3)-(-2)(2)\big) - \hat j\big(2(-3)-(-2)(1)\big) + \hat k\big(2(2)-3(1)\big)

=i^(−9+4)−j^(−6+2)+k^(4−3)=−5i^+4j^+k^= \hat i(-9+4) - \hat j(-6+2) + \hat k(4-3) = -5\hat i+4\hat j+\hat k

So the required plane has normal (−5,4,1)(-5,4,1) and passes through (1,−1,2)(1,-1,2):

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