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Question
Show that the four points A,B,C and D with position vectors 4^i+5^j+^k ,−^j−^k, 3^i+9^j+4^k and 4(−^i+^j+^k) respectively are coplanar.
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Solution
The P.V. \frac{dy}{dx} of points A,B,C and D are 4^i+5^j+^k, −^j−^k, 3^i+9^j+4^k and 4(−^i+^j+^k)
→AB=P.V. of B-P.V. of A
=−^j−^k−(4^i+5^j+^k)
=−4^i−6^j−2^k
→AC=P.V. of C-P.V. of A
=3^j+9^j+4^k−(4^i+5^j+^k)
=−^i+4^j+3^k
→AD=−4^i+4^j+4^k−(4^i+5^j+^k)
=−8^i−^j+3^k
[→AB →AC →AD]=∣∣
∣∣−4−6−2−143−8−13∣∣
∣∣
=−4(12+3)+6(−3+24)−2(1+32)
=−4(15)+6(21)−2×33
=−60+126−66=0
∴ →AB,→AC and →AD are coplanar vectors but these are co-initial vectors.
∴ Vectors →AB,→AC and →AD lie in the same plane and a
have a point in common.
∴ points A,B,C,D are coplanar.
→AB=P.V. of B-P.V. of A
=−^j−^k−(4^i+5^j+^k)
=−4^i−6^j−2^k
→AC=P.V. of C-P.V. of A
=3^j+9^j+4^k−(4^i+5^j+^k)
=−^i+4^j+3^k
→AD=−4^i+4^j+4^k−(4^i+5^j+^k)
=−8^i−^j+3^k
[→AB →AC →AD]=∣∣ ∣∣−4−6−2−143−8−13∣∣ ∣∣
=−4(12+3)+6(−3+24)−2(1+32)
=−4(15)+6(21)−2×33
=−60+126−66=0
∴ →AB,→AC and →AD are coplanar vectors but these are co-initial vectors.
∴ Vectors →AB,→AC and →AD lie in the same plane and a
have a point in common.
∴ points A,B,C,D are coplanar.
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