Question

If $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$ and $\vec{c}$ be the vector such that $\vec{a}\times\vec{c}=\vec{b}$ and $\vec{a}\cdot\vec{c}=3$, then $\vec{a}\cdot((\vec{c}\times\vec{b})-\vec{b}-\vec{c})$ is equal to

• A. 32
• B. 24
• C. 20
• D. 36

Answer 1

Answer: (B)

24

Answer 2

Given:

$\vec{a}=\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$, $\vec{a}\times \vec{c}=\vec{b}$, $\vec{a}\cdot \vec{c}=3$.

We need to find $S=\vec{a}\cdot((\vec{c}\times\vec{b})-\vec{b}-\vec{c})$.

Step 1: Write $S$ as:

$S=\underbrace{\vec{a}\cdot(\vec{c}\times\vec{b})}_{(I)} - \vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}$.

Since

$\vec{a}\cdot\vec{b}=\vec{a}\cdot(\vec{a}\times\vec{c})=0$ (dot product of a vector with its cross product is zero),

and $\vec{a}\cdot\vec{c}=3$, we get

$S=\vec{a}\cdot(\vec{c}\times\vec{b}) - 3$.

Step 2: Note that $\vec{b}=\vec{a}\times\vec{c}$. Then

$\vec{a}\cdot(\vec{c}\times\vec{b}) = \vec{a}\cdot(\vec{c}\times(\vec{a}\times\vec{c}))$.

Using the vector triple product identity:

$\vec{c}\times(\vec{a}\times\vec{c}) = (\vec{c}\cdot\vec{c})\vec{a} - (\vec{c}\cdot\vec{a})\vec{c}$,

we have

$\vec{a}\cdot\Bigl[(\vec{c}\cdot\vec{c})\vec{a} - (\vec{c}\cdot\vec{a})\vec{c}\Bigr] = (\vec{c}\cdot\vec{c})(\vec{a}\cdot\vec{a}) - (\vec{c}\cdot\vec{a})^2$.

Given:

$\vec{a}\cdot\vec{a} = 1^2+2^2+1^2=6$, $\vec{a}\cdot\vec{c}=3$.

Let $|\vec{c}|^2 = \vec{c}\cdot\vec{c}$. Then

$\vec{a}\cdot(\vec{c}\times\vec{b}) = 6|\vec{c}|^2 - 9$.

Thus,

$S=6|\vec{c}|^2 - 9 - 3 = 6|\vec{c}|^2 - 12$.

Step 3: Find $|\vec{c}|^2$ using the relation from $\vec{a}\times \vec{c}=\vec{b}$.

The magnitude of the cross product is:

$|\vec{a}\times\vec{c}|=|\vec{a}||\vec{c}|\sin\theta$,

and since $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$,

$|\vec{b}|=3\sqrt{1^2+(-1)^2+1^2}=3\sqrt{3}$.

Thus,

$|\vec{a}|\,|\vec{c}|\sin\theta=3\sqrt{3}$.

Also, from the dot product:

$\vec{a}\cdot\vec{c}=|\vec{a}||\vec{c}|\cos\theta = 3$.

Here, $|\vec{a}|=\sqrt{6}$. Let $|\vec{c}|=d$. Then:

$\cos\theta=\frac{3}{\sqrt{6}\, d}$ and $\sin\theta=\sqrt{1-\frac{9}{6d^2}}=\sqrt{\frac{6d^2-9}{6d^2}}$.

Substituting in the cross product magnitude:

$\sqrt{6}\, d \cdot \sqrt{\frac{6d^2-9}{6d^2}} = 3\sqrt{3}$.

Simplify:

$\sqrt{6d^2-9} = 3\sqrt{3} \implies 6d^2-9=27 \implies 6d^2=36 \implies d^2=6$.

Thus, $|\vec{c}|^2=6$.

Step 4: Substitute back in $S$:

$S=6\cdot6-12=36-12=24$.

Therefore, the answer is 24.