Question

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

• A. \(-1\)
• B. \(-6\)
• C. \(6\)
• D. \(1\)

Hint:

When a vector equation is equal to the zero vector, compare the coefficients of \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\) separately to form simultaneous linear equations.

Answer 1

The Correct Option is C

Step 1: Write the given vectors in component form.
We are given \[ \vec{a}=\vec{i}+2\vec{j}+3\vec{k} \] \[ \vec{b}=2\vec{i}+3\vec{j}+\vec{k} \] \[ \vec{c}=8\vec{i}+13\vec{j}+9\vec{k} \] Also, \[ x\vec{a}+y\vec{b}+z\vec{c}=0 \]

Step 2: Substitute the vectors.
Substituting the values of \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we get \[ x(\vec{i}+2\vec{j}+3\vec{k})+ y(2\vec{i}+3\vec{j}+\vec{k})+ z(8\vec{i}+13\vec{j}+9\vec{k})=0 \] Now comparing the coefficients of \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\), we get \[ x+2y+8z=0 \] \[ 2x+3y+13z=0 \] \[ 3x+y+9z=0 \]

Step 3: Solve the equations.
From the first equation, \[ x+2y+8z=0 \] we get \[ x=-2y-8z \] Substitute this in the second equation: \[ 2x+3y+13z=0 \] \[ 2(-2y-8z)+3y+13z=0 \] \[ -4y-16z+3y+13z=0 \] \[ -y-3z=0 \] Therefore, \[ y=-3z \] Now substitute \(y=-3z\) in \[ x=-2y-8z \] \[ x=-2(-3z)-8z \] \[ x=6z-8z \] \[ x=-2z \]

Step 4: Find the value of \(\dfrac{xy}{z^2}\).
Now, \[ x=-2z \] and \[ y=-3z \] Therefore, \[ xy=(-2z)(-3z) \] \[ xy=6z^2 \] Hence, \[ \frac{xy}{z^2}=\frac{6z^2}{z^2} \] \[ \frac{xy}{z^2}=6 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{6} \]