Q.The vectors from origin to the points A and B are $\vec{a}=2\hat{i}-3\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+3\hat{j}+\hat{k}$, respectively, then the area of triangle OAB is
(A) $\sqrt{340}$
(B) $\sqrt{25}$
(C) $\sqrt{229}$
(D) $\dfrac{1}{2}\sqrt{229}$

Step 1 — Find the side vectors and identify the shape.

With $A(0,4,1)$, $B(2,3,-1)$, $C(4,5,0)$, $D(2,6,2)$:

Since $\vec{AB}=\vec{DC}$, the sides $AB$ and $DC$ are equal and parallel, so $ABCD$ is a parallelogram. (This matters: for a general quadrilateral you could not use a single cross product of sides.)

Also

Step 2 — The concept: cross product as area.

For two vectors $\vec u,\vec v$ emanating from the same vertex,

So the area of $ABCD$ is $|\vec{AB}\times\vec{AD}|$.

Step 3 — Compute the cross product.

$=\hat i\big[(-1)(1)-(-2)(2)\big]-\hat j\big[(2)(1)-(-2)(2)\big]+\hat k\big[(2)(2)-(-1)(2)\big]$ …

The key here is recognising a fundamental identity that connects the dot product and cross product magnitudes. For any two vectors $\vec{a}$ and $\vec{b}$, the squared magnitude of their cross product is $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$, and the squared dot product is $|\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$, where $\theta$ is the angle between them. Adding these gives $|\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta) = |\vec{a}|^2 |\vec{b}|^2$. This is a clean, direct relation — no angle needed.

Let’s apply it step by step.

1. Write the identity For any two vectors $\vec{a}$ and $\vec{b}$,

This holds because $\sin^2 \theta + \cos^2 \theta = 1$ for any angle $\theta$.

2. Substitute the given values We are told $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144$ and $|\vec{a}| = 6$. So:

3. Solve for $|\vec{b}|$ …

Concept: |a x b| = |a| |b| sin(theta).

Given |a| = 8, |b| = 3, |a x b| = 12.

12 = 8 x 3 x sin(theta) = 24 sin(theta)

sin(theta) = 12/24 = 1/2

theta = pi/6 (30 degrees) - the principal value in [0, pi] usually quoted for the angle between two vectors when the acute solution is offered. …

The key idea is that for any vector $\vec{p}$, the magnitude of its cross product with a unit vector gives the component of $\vec{p}$ perpendicular to that unit vector. Summing these perpendicular components over all three coordinate axes yields a multiple of $|\vec{p}|^2$.

Why this works:

If $\vec{p} = p_x \hat{\imath} + p_y \hat{\jmath} + p_z \hat{k}$, then $|\vec{p} \times \hat{\imath}|$ is the magnitude of the vector perpendicular to $\hat{\imath}$, which involves only the $y$ and $z$ components. Similarly for the other axes. Adding them cleverly reconstructs the full squared magnitude.

Step-by-step:

1. Compute $|\vec{p} \times \hat{\imath}|^2$

   $\vec{p} \times \hat{\imath} = (p_y \hat{\jmath} + p_z \hat{k}) \times \hat{\imath} = p_y (\hat{\jmath} \times \hat{\imath}) + p_z (\hat{k} \times \hat{\imath}) = -p_y \hat{k} + p_z \hat{\jmath}$.

   Its magnitude squared: $|\vec{p} \times \hat{\imath}|^2 = p_y^2 + p_z^2$.

2. Compute $|\vec{p} \times \hat{\jmath}|^2$

   $\vec{p} \times \hat{\jmath} = (p_x \hat{\imath} + p_z \hat{k}) \times \hat{\jmath} = p_x (\hat{\imath} \times \hat{\jmath}) + p_z (\hat{k} \times \hat{\jmath}) = p_x \hat{k} - p_z \hat{\imath}$.

   So $|\vec{p} \times \hat{\jmath}|^2 = p_x^2 + p_z^2$.

3. Compute $|\vec{p} \times \hat{k}|^2$

   $\vec{p} \times \hat{k} = (p_x \hat{\imath} + p_y \hat{\jmath}) \times \hat{k} = p_x (\hat{\imath} \times \hat{k}) + p_y (\hat{\jmath} \times \hat{k}) = -p_x \hat{\jmath} + p_y \hat{\imath}$.

   So $|\vec{p} \times \hat{k}|^2 = p_x^2 + p_y^2$.

4. Sum the three squared magnitudes

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