Vector Algebra: JEE Main Mathematics Question with Solution
Let a,b and c be three non-zero non-coplanar vectors. Let the position vectors of four points A,B,C and D be a−b+c, λa−3b+4c, −a+2b−3c and 2a−4b+6c respectively. If AB,AC and AD are coplanar, then λ is equal to
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Hint 1 of 3
What are the components of the vectors AB, AC, and AD in terms of the basis vectors a,b,c?
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Correct answer
The vectors AB,AC and AD are coplanar when the scalar triple product [ABACAD]=0, which yields λ=2.
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Solution
StepWorking
01given
Points A,B,C,D have position vectors relative to an origin:
OA=a−b+cOB=λa−3b+4cOC=−a+2b−3cOD=2a−4b+6c
where a,b,c are non-zero, non-coplanar vectors, and AB,AC,AD are coplanar.
02goal
Find the value of scalar λ for which AB,AC,AD are coplanar.
03approach
First compute the displacement vectors AB=OB−OA, AC=OC−OA, and AD=OD−OA as linear combinations of the basis {a,b,c}. Since the basis vectors are non-coplanar ([abc]=0), the coplanarity condition [ABACAD]=0 is equivalent to setting the determinant of their coordinate coefficients to zero.
04execute
Compute the displacement vectors:
AB=(λ−1)a−2b+3cAC=−2a+3b−4cAD=a−3b+5c
05execute
Evaluate the scalar triple product determinant and set to zero:
λ−1−21−23−33−45=0
Expanding along the first row:
(λ−1)(15−12)−(−2)(−10−(−4))+3(6−3)=03(λ−1)+2(−6)+3(3)=03λ−3−12+9=03λ−6=0⟹λ=2.
✓verify
Substitute λ=2: the first row becomes (1−23).
Notice that Row 1 + Row 2 = (1−2−2+33−4)=(−11−1), and Row 3 + 2*(Row 1 + Row 2) = (1−35)+(−22−2)=(−1−13).
Direct determinant check: 1(3)−(−2)(−6)+3(3)=3−12+9=0. Holds identically.
Hints that build this answer step by step
What are the components of the vectors AB, AC, and AD in terms of the basis vectors a,b,c?
AB=(λ−1)a−2b+3c, AC=−2a+3b−4c, AD=a−3b+5c
Since a,b,c are non-coplanar, what condition ensures that AB,AC, and AD are coplanar?
The determinant formed by the coefficients of AB, AC, and AD must equal zero.
Evaluating the determinant λ−1−21−23−33−45=0, what is the value of λ?
Why can we set the determinant of coefficients directly to zero instead of taking the cross product and dot product with a,b,c?
Because [ABACAD]=det(M)[abc], where M is the matrix of coordinates. Since a,b,c are non-coplanar, [abc]=0, so the scalar triple product vanishes if and only if det(M)=0.