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Question : 64 of 160
Marks:
+1,
-0
Solution:
Let the given three point be
A≡(0,1,2),B≡(3,0,2)‌ and ‌C≡(4,5,0)Equation of plane passing through three points is
|| x−x1 | y−y1 | z−z1 |
| x2−x1 | y2−y1 | z2−z1 |
| x3−x1 | y3−y1 | z3−z1 |
|=0‌⇒‌‌|| x−0 | y−1 | z−2 |
| 3−0 | 0−1 | 2−2 |
| 4−0 | 5−1 | 0−2 |
|=0‌⇒‌‌||=0 ⇒(z−2)[(3)(4)−(4)(−1)]−0+(−2)[‌[(x)(−1)‌−3(y−1)]=0
‌⇒‌‌(z−2)[12+4]−2[−x−3y+3]=0
‌⇒‌‌16z−32+2x+6y−6=0‌⇒‌‌2x+6y+16z=38⇒x+3y+8z=19
⇒‌=‌‌‌x+‌y+‌z=‌ Comparing to
lx+my+nz=p, we have
l=‌,m=‌‌ and ‌n=‌∴‌‌|l|+|m|+|n|=‌=‌
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