Find the Value of Vector x

To find the value of $x$, we use the given information from the problem stem:

Given:

1. $\vec{OP} = x\hat{i} + y\hat{j} - \hat{k}$ and $\vec{OQ} = -\hat{i} + 2\hat{j} + x\hat{k}$, where $x, y \in \mathbb{R}$ and $x > 0$.
2. $\vec{OP}$ is perpendicular to $\vec{OQ}$, i.e., $\vec{OP} \cdot \vec{OQ} = 0$.
3. The magnitude of the cross product is $|\vec{OP} \times \vec{OQ}| = 3\sqrt{2}$.

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Step 1: Apply the Condition for Perpendicularity

Since $\vec{OP} \perp \vec{OQ}$, their dot product is zero: $\vec{OP} \cdot \vec{OQ} = (x)(-1) + (y)(2) + (-1)(x) = 0$ $-x + 2y - x = 0$ $2y - 2x = 0 \implies y = x$

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Step 2: Relate Cross Product Magnitude to Vectors' Magnitudes

Since the angle between $\vec{OP}$ and $\vec{OQ}$ is $90^\circ$, we have: $|\vec{OP} \times \vec{OQ}| = |\vec{OP}| |\vec{OQ}| \sin 90^\circ = |\vec{OP}| |\vec{OQ}|$

Squaring both sides gives: $|\vec{OP}|^2 |\vec{OQ}|^2 = |\vec{OP} \times \vec{OQ}|^2 = (3\sqrt{2})^2 = 18$

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Step 3: Express the Magnitudes in terms of $x$

Using $y = x$: $|\vec{OP}|^2 = x^2 + y^2 + (-1)^2 = x^2 + x^2 + 1 = 2x^2 + 1$ $|\vec{OQ}|^2 = (-1)^2 + 2^2 + x^2 = 1 + 4 + x^2 = x^2 + 5$

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Step 4: Solve for $x$

Substitute the expressions for $|\vec{OP}|^2$ and $|\vec{OQ}|^2$ into the squared magnitude equation: $(2x^2 + 1)(x^2 + 5) = 18$ $2x^4 + 10x^2 + x^2 + 5 = 18$ $2x^4 + 11x^2 - 13 = 0$

Factorizing the quadratic equation in $x^2$: $2x^4 - 2x^2 + 13x^2 - 13 = 0$ $2x^2(x^2 - 1) + 13(x^2 - 1) = 0$ $(2x^2 + 13)(x^2 - 1) = 0$

Since $x \in \mathbb{R}$, $2x^2 + 13 > 0$ for all real $x$. Therefore, we must have: $x^2 - 1 = 0 \implies x^2 = 1$

Given that $x > 0$: $x = 1$

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Final Answer:

The value of $x$ is 1.