Self Dot Product

Using

$$|\vec a+\vec b+\vec c|^2 = |\vec a|^2+|\vec b|^2+|\vec c|^2 +2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a),$$

and comparing with the given condition,

$$2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0$$

Hence,

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

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. $$\frac{27}{2}$$

Step 1: Translate the Perpendicularity Conditions into Equations

We are given that each vector is perpendicular to the sum of the remaining two vectors:

$$\vec{p} \cdot (\vec{q} + \vec{r}) = 0 \implies \vec{p} \cdot \vec{q} + \vec{p} \cdot \vec{r} = 0$$

$$\vec{q} \cdot (\vec{p} + \vec{r}) = 0 \implies \vec{q} \cdot \vec{p} + \vec{q} \cdot \vec{r} = 0$$

$$\vec{r} \cdot (\vec{p} + \vec{q}) = 0 \implies \vec{r} \cdot \vec{p} + \vec{r} \cdot \vec{q} = 0$$

Step 2: Sum the Equations

Add the three scalar equations together:

$$(\vec{p} \cdot \vec{q} + \vec{p} \cdot \vec{r}) + (\vec{q} \cdot \vec{p} + \vec{q} \cdot \vec{r}) + (\vec{r} \cdot \vec{p} + \vec{r} \cdot \vec{q}) = 0$$

Since the dot product is commutative ($$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$$), we can collect like terms:

$$2(\vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{r} + \vec{r} \cdot \vec{p}) = 0$$

Dividing by 2 yields:

$$\vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{r} + \vec{r} \cdot \vec{p} = 0$$

Step 3: Use the Vector Magnitude Identity

We expand the square of the magnitude of the sum vector using the algebraic identity for vectors:

$$|\vec{p} + \vec{q} + \vec{r}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + |\vec{r}|^2 + 2(\vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{r} + \vec{r} \cdot \vec{p})$$

Step 4: Substitute Known Values

We are given the individual magnitudes:

$$|\vec{p}| = 2 \implies |\vec{p}|^2 = 4$$

$$|\vec{q}| = 3 \implies |\vec{q}|^2 = 9$$

$$|\vec{r}| = 4 \implies |\vec{r}|^2 = 16$$

From Step 2, we know that the cross-product/dot-product combination term is zero:

$$2(\vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{r} + \vec{r} \cdot \vec{p}) = 0$$

Substitute these values into the expansion equation:

$$|\vec{p} + \vec{q} + \vec{r}|^2 = 4 + 9 + 16 + 0$$

$$|\vec{p} + \vec{q} + \vec{r}|^2 = 29$$

Step 5: Solve for the Final Magnitude

Taking the square root of both sides gives the absolute magnitude:

$$|\vec{p} + \vec{q} + \vec{r}| = \sqrt{29}$$