Q.Find $|\vec{a} \times \vec{b}|$, if $\vec{a}=\hat{i}-7\hat{j}+7\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}$.

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

5. Multiply by 2 as per the given expression The original expression is $2\left\{|\vec{p} \times \hat{\imath}|^2+|\vec{p} \times \hat{\jmath}|^2+|\vec{p} \times \hat{k}|^2\right\}$. Substituting the sum: $2 \times (2|\vec{p}|^2) = 4|\vec{p}|^2$.

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.

Step 4 — Take the magnitude.

Cross-check with the diagonal formula. For a parallelogram, area $=\dfrac12|\vec{AC}\times\vec{BD}|$. Here $\vec{AC}=(4,1,-1)$ and $\vec{BD}=(0,3,3)$, giving $\vec{AC}\times\vec{BD}=(6,-12,12)$ with magnitude $\sqrt{36+144+144}=18$, and $\tfrac12(18)=9$. ✓ Same answer.

ANSWER: A

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 supplementary value 5pi/6 also satisfies sin theta = 1/2 but is not among the options; pi/6 is.)

ANSWER: B

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 $|\vec{b}|$

(Magnitude is positive, so we take the positive root.)