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Vector Algebra question

2025 · Shift 2 · Q28
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Vector Algebra question

2025 · Shift 2 · Q28

JEE AdvancedMathematicsVector AlgebraNumerical+4 / −1
Consider the vectors x⃗=ı^+2ȷ^+3k^,y⃗=2ı^+3ȷ^+k^, and z⃗=3ı^+ȷ^+2k^\vec{x}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}, \quad \vec{y}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}, \quad \text { and } \quad \vec{z}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k}x=^+2^​+3k^,y​=2^+3^​+k^, and z=3^+^​+2k^ For two distinct positive real numbers α\alphaα and β\betaβ, define X⃗=αx⃗+βy⃗−z⃗,Y⃗=αy⃗+βz⃗−x⃗, and Z⃗=αz⃗+βx⃗−y⃗.\vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}, \quad \vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}, \quad \text { and } \quad \vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y} .X=αx+βy​−z,Y=αy​+βz−x, and Z=αz+βx−y​. If the vectors X⃗,Y⃗\vec{X}, \vec{Y}X,Y, and Z⃗\vec{Z}Z lie in a plane, then the value of α+β−3\alpha+\beta-3α+β−3 is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: -2

  1. Given vectors

x⃗=(1,2,3),y⃗=(2,3,1),z⃗=(3,1,2).\vec{x}=(1,2,3),\quad \vec{y}=(2,3,1),\quad \vec{z}=(3,1,2).x=(1,2,3),y​=(2,3,1),z=(3,1,2).

We define

X⃗=αx⃗+βy⃗−z⃗,\vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z},X=αx+βy​−z, Y⃗=αy⃗+βz⃗−x⃗,\vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x},Y=αy​+βz−x, Z⃗=αz⃗+βx⃗−y⃗.\vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y}.Z=αz+βx−y​.

Since X⃗,Y⃗,Z⃗\vec{X},\vec{Y},\vec{Z}X,Y,Z lie in a plane through the origin, they must be linearly dependent. Hence

[X⃗ Y⃗ Z⃗]=0,[\vec{X}\ \vec{Y}\ \vec{Z}] = 0,[X Y Z]=0,

i.e. the determinant of the matrix having X⃗,Y⃗,Z⃗\vec{X},\vec{Y},\vec{Z}X,Y,Z as columns is zero.

  1. Compute X⃗,Y⃗,Z⃗\vec{X},\vec{Y},\vec{Z}X,Y,Z component-wise

First,

X⃗=α(1,2,3)+β(2,3,1)−(3,1,2).\vec{X}=\alpha(1,2,3)+\beta(2,3,1)-(3,1,2).X=α(1,2,3)+β(2,3,1)−(3,1,2).

So,

X⃗=(α+2β−3, 2α+3β−1, 3α+β−2).\vec{X}=(\alpha+2\beta-3,\ 2\alpha+3\beta-1,\ 3\alpha+\beta-2).X=(α+2β−3, 2α+3β−1, 3α+β−2).

Similarly,

Y⃗=α(2,3,1)+β(3,1,2)−(1,2,3),\vec{Y}=\alpha(2,3,1)+\beta(3,1,2)-(1,2,3),Y=α(2,3,1)+β(3,1,2)−(1,2,3),

hence

Y⃗=(2α+3β−1, 3α+β−2, α+2β−3).\vec{Y}=(2\alpha+3\beta-1,\ 3\alpha+\beta-2,\ \alpha+2\beta-3).Y=(2α+3β−1, 3α+β−2, α+2β−3).

And,

Z⃗=α(3,1,2)+β(1,2,3)−(2,3,1),\vec{Z}=\alpha(3,1,2)+\beta(1,2,3)-(2,3,1),Z=α(3,1,2)+β(1,2,3)−(2,3,1),

thus

Z⃗=(3α+β−2, α+2β−3, 2α+3β−1).\vec{Z}=(3\alpha+\beta-2,\ \alpha+2\beta-3,\ 2\alpha+3\beta-1).Z=(3α+β−2, α+2β−3, 2α+3β−1).

  1. Observe the cyclic form

Let

A=α+2β−3,B=2α+3β−1,C=3α+β−2.A=\alpha+2\beta-3,\quad B=2\alpha+3\beta-1,\quad C=3\alpha+\beta-2.A=α+2β−3,B=2α+3β−1,C=3α+β−2.

Then

X⃗=(A,B,C),Y⃗=(B,C,A),Z⃗=(C,A,B).\vec{X}=(A,B,C),\quad \vec{Y}=(B,C,A),\quad \vec{Z}=(C,A,B).X=(A,B,C),Y=(B,C,A),Z=(C,A,B).

So the determinant is

A & B & C\\ B & C & A\\ C & A & B \end{vmatrix}=0.$$ Using the standard identity, $$\begin{vmatrix} A & B & C\\ B & C & A\\ C & A & B \end{vmatrix} =-(A+B+C)\left(A^2+B^2+C^2-AB-BC-CA\right).$$ Therefore, $$(A+B+C)\left(A^2+B^2+C^2-AB-BC-CA\right)=0.$$ Now, $$A^2+B^2+C^2-AB-BC-CA=\frac12\left[(A-B)^2+(B-C)^2+(C-A)^2\right].$$ This is zero iff $A=B=C$. --- 4. **Use the condition that $\alpha,\beta$ are distinct** Check when $A=B=C$: $$A=B \implies \alpha+2\beta-3=2\alpha+3\beta-1 \implies \alpha+\beta=-2,$$ which is impossible for positive real numbers. So the second factor cannot be zero. Hence we must have $$A+B+C=0.$$ Now, $$A+B+C=(\alpha+2\beta-3)+(2\alpha+3\beta-1)+(3\alpha+\beta-2).$$ So, $$A+B+C=6\alpha+6\beta-6=6(\alpha+\beta-1).$$ Thus, $$\alpha+\beta-1=0 \implies \alpha+\beta=1.$$ Therefore, $$\alpha+\beta-3=1-3=-2.$$ --- 5. **Final answer** $$\boxed{-2}$$
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