Dot Product in Component Form

$$M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}$$

$$\vec{u} = \alpha\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \gamma\hat{k}$$

$$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \beta\hat{j} + \delta\hat{k}$$

$$\vec{w} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \mu\hat{k}$$

$$\alpha^2 + \frac{1}{3} + \gamma^2 = 1 \quad \therefore \alpha^2 + \gamma^2 = \frac{2}{3} \quad \dots(1)$$

$$\frac{1}{2} + \beta^2 + \delta^2 = 1 \quad \beta^2 + \delta^2 = \frac{1}{2} \quad \dots(2)$$

$$\frac{1}{2} + \frac{1}{3} + \mu^2 = 1 \quad \mu^2 = \frac{1}{6} \quad \dots(3)$$

$$\gamma^2 + \delta^2 + \mu^2 = 1$$

$$\gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6}$$

For Q:

$$x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$$

$$x\vec{u} \cdot \vec{u} + y\vec{v} \cdot \vec{u} + z\vec{w} \cdot \vec{u} = \vec{u} \cdot \hat{j}$$

$$x = \frac{1}{\sqrt{3}}$$

For R: 

$$\vec{u}, \vec{v}, \vec{w}$$ are mutually perpendicular unit vectors.

$$\vert{}\vec{u} \cdot (\vec{v} \times \vec{w})\vert{}$$

$$\begin{bmatrix} \vec{u} & \vec{v} & \vec{w} \end{bmatrix} = 1$$

For S: 

$$\vec{u} \times (\vec{v} \times \vec{w})$$

$$(\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$$

$$= 0 - 0 = 0$$

Solution :

Given,

$$\vec a\times\vec c=\vec b$$

Take dot product of $$\vec a$$ on both sides :

$$\vec a\cdot(\vec a\times\vec c)=\vec a\cdot\vec b$$

But,

$$\vec a\cdot(\vec a\times\vec c)=0$$

since a vector is always perpendicular to its cross product.

Therefore,

$$\vec a\cdot\vec b=0$$

Now,

$$\vec a=\hat i+\hat j+\hat k$$

and

$$\vec b=\hat i-\hat j+\hat k$$

Hence,

$$\vec a\cdot\vec b=(1)(1)+(1)(-1)+(1)(1)$$

$$=1-1+1$$

$$=1$$

But,

$$\vec a\cdot\vec b=1\neq 0$$

which contradicts the necessary condition

$$\vec a\cdot\vec b=0$$

for the existence of a vector $$\vec c$$ satisfying

$$\vec a\times\vec c=\vec b$$

Therefore, no such vector $$\vec c$$ exists.

Since $$\vec{d}$$ is coplanar with $$\vec{a}$$ and $$\vec{b}$$, it can be written as a linear combination of them:

$$\vec{d} = \lambda \vec{a} + \mu \vec{b}$$

Substitute the given vectors $$\vec{a}$$ and $$\vec{b}$$:

$$\vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$$ $$\vec{d} = (\lambda + \mu)\hat{i} + \lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}$$

We are given that $$\vec{d}$$ is perpendicular to $$\vec{a}$$ ($$\vec{d} \cdot \vec{a} = 0$$):

$$(\lambda \vec{a} + \mu \vec{b}) \cdot \vec{a} = 0$$ $$\lambda |\vec{a}|^2 + \mu (\vec{b} \cdot \vec{a}) = 0$$

Let's calculate the required dot products:

• $$|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$$
• $$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(1) + (1)(1) = 1 - 1 + 1 = 1$$

Substitute these values back into the condition:

$$3\lambda + \mu(1) = 0 \implies \mu = -3\lambda$$

Now, update our expression for $$\vec{d}$$ in terms of $$\lambda$$ only:

$$\vec{d} = \lambda \vec{a} - 3\lambda \vec{b} = \lambda (\vec{a} - 3\vec{b})$$

Let's compute $$\vec{a} - 3\vec{b}$$:

$$\vec{a} - 3\vec{b} = (\hat{i} + \hat{j} + \hat{k}) - 3(\hat{i} - \hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$$

So, the vector $$\vec{d}$$ simplifies to:

$$\vec{d} = \lambda(-2\hat{i} + 4\hat{j} - 2\hat{k})$$

We are given $$\vec{d} \cdot \vec{c} = 6$$, where $$\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$$:

$$\lambda(-2\hat{i} + 4\hat{j} - 2\hat{k}) \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 6$$ $$\lambda \left[ (-2)(1) + (4)(2) + (-2)(-1) \right] = 6$$ $$\lambda \left[ -2 + 8 + 2 \right] = 6$$ $$8\lambda = 6 \implies \lambda = \frac{3}{4}$$

Now we can write the exact vector $$\vec{d}$$:

$$\vec{d} = \frac{3}{4}(-2\hat{i} + 4\hat{j} - 2\hat{k}) = -\frac{3}{2}\hat{i} + 3\hat{j} - \frac{3}{2}\hat{k}$$

Finally, compute the square of its magnitude:

$$|\vec{d}|^2 = \left(-\frac{3}{2}\right)^2 + (3)^2 + \left(-\frac{3}{2}\right)^2$$ $$|\vec{d}|^2 = \frac{9}{4} + 9 + \frac{9}{4}$$ $$|\vec{d}|^2 = \frac{18}{4} + 9 = \frac{9}{2} + 9 = \frac{27}{2}$$

Correct Answer: C. $$\frac{27}{2}$$