Question

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}$ and $\vec{c}=\hat{j}-\hat{k}$ be three vectors such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$. If the length of projection vector of the vector $\vec{b}$ on the vector $\vec{a} \times \vec{c}$ is $l$, then the value of $3l^2$ is equal to _________.

Hint:

Remember the cyclic property of the scalar triple product: $[\vec{a} \ \vec{b} \ \vec{c}] = [\vec{b} \ \vec{c} \ \vec{a}] = [\vec{c} \ \vec{a} \ \vec{b}]$. Swapping any two vectors negates the value, e.g., $[\vec{a} \ \vec{b} \ \vec{c}] = -[\vec{b} \ \vec{a} \ \vec{c}]$. This can simplify calculations significantly.

Answer 1

Correct Answer: 2

The length of the projection of a vector \(\vec u\) on a vector \(\vec v\) is \[ \text{Projection length}=\frac{|\vec u\cdot\vec v|}{|\vec v|} \] Here, \[ l=\frac{|\vec b\cdot(\vec a\times\vec c)|}{|\vec a\times\vec c|} \] Step 1: Evaluate the numerator \[ \vec b\cdot(\vec a\times\vec c) \] is a scalar triple product. Using the cyclic property, \[ \vec b\cdot(\vec a\times\vec c)=\vec c\cdot(\vec b\times\vec a) \] Since \[ \vec a\times\vec b=\vec c \quad\Rightarrow\quad \vec b\times\vec a=-\vec c \] \[ \vec b\cdot(\vec a\times\vec c) =\vec c\cdot(-\vec c) =-|\vec c|^2 \] Now, \[ \vec c=\hat j-\hat k \quad\Rightarrow\quad |\vec c|^2=1^2+(-1)^2=2 \] Hence, \[ |\vec b\cdot(\vec a\times\vec c)|=2 \] Step 2: Evaluate the denominator 

\[ \vec a\times\vec c=-2\hat i+\hat j+\hat k \] \[ |\vec a\times\vec c| =\sqrt{(-2)^2+1^2+1^2} =\sqrt6 \] Step 3: Find \(l\) \[ l=\frac{2}{\sqrt6} \] Step 4: Compute \(3l^2\) \[ 3l^2 =3\left(\frac{2}{\sqrt6}\right)^2 =3\cdot\frac{4}{6} =2 \] Answer: \(\boxed{2}\)