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Question : 25 of 90
Marks:
+1,
-0
Solution:
‌×=×5‌⇒×(−)=0‌|rl(−)‌∴=λ(−)‌(6,9,12)=λ[x−α,y−11,z+2]‌‌=‌=‌‌4y−44=3z+6‌4y−3z=50‌6x+9y+12z=−12‌2x+3y+4z=−4‌(∵x−2y+z=5)‌2x−4y+2z=10‌+‌‌−‌7y+2z=−14‌‌...(2)‌8y−6z=100‌21y+6z=−42‌29y=58‌y=2,z=−14‌∴x−4−14=5‌x=23‌c=(23,2,−14)‌c⋅(1,1,1)=23+2−14=11‌
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