Question

If \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \), \( \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k} \), \( \vec{c} = 8\hat{i} + 13\hat{j} + 9\hat{k} \) and \( x\vec{a} + y\vec{b} + z\vec{c} = \vec{0} \), then \( \frac{xy}{z^2} = \)

• A. -1
• B. -6
• C. 6
• D. 1

Hint:

If \(x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}\), equate the components (i, j, k) to zero to form a system of linear equations in x, y, z.
Solve the system to find the ratios between x, y, and z (e.g., express x and y in terms of z).
Substitute these relations into the expression to be evaluated.
If a non-trivial solution exists for x,y,z, the vectors \(\vec{a},\vec{b},\vec{c}\) are linearly dependent (coplanar if 3D).

Answer 1

The Correct Option is C

Given \( x\vec{a} + y\vec{b} + z\vec{c} = \vec{0} \). This means the vectors \(\vec{a}, \vec{b}, \vec{c}\) are linearly dependent if \(x,y,z\) are not all zero. If \(\vec{a}, \vec{b}, \vec{c}\) are coplanar, their scalar triple product is zero, or one can be written as a linear combination of the others. The equation is: \( x(\hat{i} + 2\hat{j} + 3\hat{k}) + y(2\hat{i} + 3\hat{j} + \hat{k}) + z(8\hat{i} + 13\hat{j} + 9\hat{k}) = 0\hat{i} + 0\hat{j} + 0\hat{k} \) Equating coefficients of \(\hat{i}, \hat{j}, \hat{k}\): For \(\hat{i}\): \(x + 2y + 8z = 0\) --- (1) For \(\hat{j}\): \(2x + 3y + 13z = 0\) --- (2) For \(\hat{k}\): \(3x + y + 9z = 0\) --- (3) This is a homogeneous system of linear equations for \(x,y,z\). For a non-trivial solution (other than \(x=y=z=0\)), the determinant of the coefficient matrix must be zero (meaning vectors are coplanar). Let's try to find a relationship between \(x,y,z\). From (3), \(y = -3x - 9z\). Substitute into (1): \(x + 2(-3x - 9z) + 8z = 0\) \(x - 6x - 18z + 8z = 0\) \(-5x - 10z = 0 \Rightarrow -5x = 10z \Rightarrow x = -2z\). Now substitute \(x=-2z\) into the expression for y: \(y = -3(-2z) - 9z = 6z - 9z = -3z\). So we have \(x = -2z\) and \(y = -3z\). Let's check if these satisfy equation (2): \(2x + 3y + 13z = 2(-2z) + 3(-3z) + 13z = -4z - 9z + 13z = -13z + 13z = 0\). Yes, it is satisfied. So, for any non-zero \(z\), we have \(x=-2z\) and \(y=-3z\). We need to calculate \( \frac{xy}{z^2} \). \( \frac{xy}{z^2} = \frac{(-2z)(-3z)}{z^2} = \frac{6z^2}{z^2} \). Assuming \(z \neq 0\) (for a non-trivial linear combination), we can cancel \(z^2\). \( \frac{xy}{z^2} = 6 \). This matches option (c). \[ \boxed{6} \]