1
Question
The position vector of the points A,B,C,D are 3^i−2^j−^k,2^i+3^j−4^k,−^i+^j+2^k,4^i+5^j+γ^k respectively. If the points A,B,C,D lie on a plane the value of γ is
Open in App
Solution
If the points are coplanar then their position vectors →a,→b,→c,→d must satisfy
[→b−→a,→c−→a,→d−→a]=0
Let →a=3^i−2^j−^k,→b=2^i+3^j−4^k,→c=−^i+^j+2^k,→d=4^i+5^j+γ^k
→b−→a=(2^i+3^j−4^k)−(3^i−2^j−^k)
=−^i+5^j−3^k
→c−→a=(→b−→a)−(3^i−2^j−^k)
=−4^i+3^j+3^k
→d−→a=(4^i+5^j+γ^k)−(3^i−2^j−^k)
=^i+7^j+(1+γ)^k
Now [→b−→a,→c−→a,→d−→a]
=∣∣
∣∣−15−3−43317γ+1∣∣
∣∣
=−1(3γ+3−21)−5(−4γ−4−3)−3(−28−3)
=−1(3γ−18)−5(−4γ−7)−3(−31)
⇒146+17γ=0 on simplification
∴γ=−14617
[→b−→a,→c−→a,→d−→a]=0
Let →a=3^i−2^j−^k,→b=2^i+3^j−4^k,→c=−^i+^j+2^k,→d=4^i+5^j+γ^k
→b−→a=(2^i+3^j−4^k)−(3^i−2^j−^k)
=−^i+5^j−3^k
→c−→a=(→b−→a)−(3^i−2^j−^k)
=−4^i+3^j+3^k
→d−→a=(4^i+5^j+γ^k)−(3^i−2^j−^k)
=^i+7^j+(1+γ)^k
Now [→b−→a,→c−→a,→d−→a]
=∣∣ ∣∣−15−3−43317γ+1∣∣ ∣∣
=−1(3γ+3−21)−5(−4γ−4−3)−3(−28−3)
=−1(3γ−18)−5(−4γ−7)−3(−31)
⇒146+17γ=0 on simplification
∴γ=−14617
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
