JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023If the four points, whose position vectors are 3i^−4j^+2k^, i^+2j^−k^, −2i^−j^+3k^, and 5i^−2αj^+4k^ are coplanar, then α is equal to:
Choose the correct answer:
- A.
1773
(Correct Answer) - B.
17107
- C.
17−73
1773
Explanation
Solving
Let the position vectors be A,B,C,D.
A=(3,−4,2),B=(1,2,−1),C=(−2,−1,3),D=(5,−2α,4)
Calculate vectors AB,AC,AD:
AB=B−A=(1−3,2+4,−1−2)=(−2,6,−3)
AC=C−A=(−2−3,−1+4,3−2)=(−5,3,1)
AD=D−A=(5−3,−2α+4,4−2)=(2,4−2α,2)
For coplanarity, the scalar triple product is zero:
−2−52amp;6amp;3amp;4−2αamp;−3amp;1amp;2=0
Expanding along the first row:
−2[6−(4−2α)]−6[−10−2]−3[−5(4−2α)−6]=0
−2[2+2α]−6[−12]−3[−20+10α−6]=0
−4−4α+72+60−30α+18=0
−34α+146=0
34α=146
α=34146=1773
Correct Option: (1)
Explanation
Solving
Let the position vectors be A,B,C,D.
A=(3,−4,2),B=(1,2,−1),C=(−2,−1,3),D=(5,−2α,4)
Calculate vectors AB,AC,AD:
AB=B−A=(1−3,2+4,−1−2)=(−2,6,−3)
AC=C−A=(−2−3,−1+4,3−2)=(−5,3,1)
AD=D−A=(5−3,−2α+4,4−2)=(2,4−2α,2)
For coplanarity, the scalar triple product is zero:
−2−52amp;6amp;3amp;4−2αamp;−3amp;1amp;2=0
Expanding along the first row:
−2[6−(4−2α)]−6[−10−2]−3[−5(4−2α)−6]=0
−2[2+2α]−6[−12]−3[−20+10α−6]=0
−4−4α+72+60−30α+18=0
−34α+146=0
34α=146
α=34146=1773
Correct Option: (1)