Question:

If \( \mathbf{a} = (2x + y)\mathbf{i} + 3\mathbf{j} + 9\mathbf{k} \) and \( \mathbf{b} = 2\mathbf{i} + \mathbf{j} - (x - y)\mathbf{k} \) are two collinear vectors, then \( x^3 + 27y^3 = \)

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For two collinear vectors \( \mathbf{a} \) and \( \mathbf{b} \), \( \mathbf{a} = \lambda \mathbf{b} \) for some scalar \( \lambda \). Equate the corresponding components of the vectors to form a system of equations and solve for \( x \) and \( y \). Finally, substitute the values of \( x \) and \( y \) into the expression \( x^3 + 27y^3 \).
Updated On: May 12, 2025
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The Correct Option is D

Solution and Explanation

Since vectors \( \mathbf{a} \) and \( \mathbf{b} \) are collinear, there exists a scalar \( \lambda \) such that \( \mathbf{a} = \lambda \mathbf{b} \).
$$ (2x + y)\mathbf{i} + 3\mathbf{j} + 9\mathbf{k} = \lambda (2\mathbf{i} + \mathbf{j} - (x - y)\mathbf{k}) $$ $$ (2x + y)\mathbf{i} + 3\mathbf{j} + 9\mathbf{k} = 2\lambda \mathbf{i} + \lambda \mathbf{j} - \lambda(x - y)\mathbf{k} $$ Equating the coefficients of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \): $$ 2x + y = 2\lambda \quad \cdots (1) $$ $$ 3 = \lambda \quad \cdots (2) $$ $$ 9 = -\lambda(x - y) \quad \cdots (3) $$ From equation (2), \( \lambda = 3 \).
Substitute \( \lambda = 3 \) into equation (1): $$ 2x + y = 2(3) = 6 \quad \cdots (4) $$ Substitute \( \lambda = 3 \) into equation (3): $$ 9 = -3(x - y) $$ $$ -3 = x - y $$ $$ y - x = 3 \quad \cdots (5) $$ Now we have a system of two linear equations with two variables \( x \) and \( y \): $$ 2x + y = 6 $$ $$ -x + y = 3 $$ Subtracting the second equation from the first: $$ (2x + y) - (-x + y) = 6 - 3 $$ $$ 3x = 3 \implies x = 1 $$ Substitute \( x = 1 \) into \( y - x = 3 \): $$ y - 1 = 3 \implies y = 4 $$ Now we need to find the value of \( x^3 + 27y^3 \): $$ x^3 + 27y^3 = (1)^3 + 27(4)^3 = 1 + 27(64) = 1 + 1728 = 1729 $$
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