Question

If \(\vec{a} = x\hat{i} + x\hat{j} + y\hat{k}\), \(\vec{b} = \hat{i} + \hat{k}\) and \(\vec{c} = y\hat{i} + y\hat{j} + z\hat{k}\) are three vectors such that \(\vec{a} \times \vec{b} \perp \vec{c}\), then

• A. \(y^2 = xz\)
• B. \(x^2 = yz\)
• C. \(z^2 = xy\)
• D. \(xyz = 1\)

Hint:

Use \((\vec{a}\times\vec{b})\cdot\vec{c}=0\) for perpendicular vectors.

Answer 1

The Correct Option is A

Step 1: Idea.
Two vectors are perpendicular when their dot product is zero. So we need \((\vec{a} \times \vec{b}) \cdot \vec{c} = 0\).

Step 2: Find the cross product.
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & x & y \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(x - 0) - \hat{j}(x - y) + \hat{k}(0 - x) \]
So \(\vec{a} \times \vec{b} = x\hat{i} + (y - x)\hat{j} - x\hat{k}\).

Step 3: Take the dot product with c.
\[ x(y) + (y - x)(y) - x(z) = xy + y^2 - xy - xz = y^2 - xz \]
Setting this equal to zero gives \(y^2 = xz\).

Step 4: Check the options.
Option 1 is \(y^2 = xz\), which matches. Options 2, 3 and 4 do not follow from the condition, since the terms \(x^2\), \(z^2\) and \(xyz\) never appear in our result.

Final Answer:
The condition is \(y^2 = xz\), which is option 1. \[ \boxed{y^2 = xz} \]