Question

find the equition of hyperplan passing through p=(1,2,3) q=(-2,-1,3) r=(2,-3,1)​

Answer 1

$ANSWER</p><p></p><p>Vector equation of a plane passing through three points with position vectors </p><p></p><p></p><p>a,b,c is </p><p></p><p></p><p>(r−a).[(b−a)×(c−a)]=0</p><p></p><p></p><p>Now, the plane passes through the points</p><p></p><p></p><p>A(2,3,4),B(−3,5,1) and C(4,−1,2)</p><p></p><p></p><p>b−a=−3i^+5j^+k^−2i^−3j^−4k^=−5i^+2j^−3k^</p><p></p><p></p><p>c−a=4i^−j^+2k^−2i^−3j^−4k^=2i^−4j^−2k^</p><p></p><p></p><p>(b−a)×(c−a)=∣∣∣∣∣∣∣∣i^−52j^2−4k^−3−2∣∣∣∣∣∣∣∣</p><p></p><p></p><p>=(−4−12)i^−(10+6)j^+(20−4)k^</p><p></p><p></p><p>=−16i^−16j^+16k^</p><p></p><p></p><p>∴vector equation of plane is (xi^+yj^+zk^−2i^−3j^−4k^)(8i^−16j^−16k^)=0</p><p></p><p></p><p>⇒[(x−2)i^+(y−3)j^+(z−4)k^](8i^−16j^−16k^)=0</p><p></p><p></p><p>⇒8x−16−16y+48−16z+64=0</p><p></p><p></p><p>⇒8x−16y−16z+96=0</p><p></p><p></p><p>or</p><p></p><p> x−2y−2z+12=0 is the required equation of the plane.</p><p></p><p>$

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