Question

If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{p}, \vec{q}, \vec{r}$ are vectors defined by:}\\ \[ \vec{p} = \frac{\vec{a} \times \vec{c}}{[\vec{a} \, \vec{b} \, \vec{c}]}, \quad \vec{q} = \frac{\vec{c} \times \vec{b}}{[\vec{a} \, \vec{b} \, \vec{c}]}, \quad \vec{r} = \frac{\vec{b} \times \vec{a}}{[\vec{a} \, \vec{b} \, \vec{c}]}, \] then $(\vec{i} + \vec{b}) \cdot \vec{p} + (\vec{b} + \vec{c}) \cdot \vec{q} + (\vec{c} + \vec{a}) \cdot \vec{r}$ is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is D

1. Understand the vectors:

Given non-coplanar vectors \( \bar{a}, \bar{b}, \bar{c} \), and the vectors \( \bar{p}, \bar{q}, \bar{r} \) defined as:

\[ \bar{p} = \frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \quad \bar{q} = \frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \quad \bar{r} = \frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]} \]

where \( [\bar{a} \bar{b} \bar{c}] \) is the scalar triple product.

2. Compute each dot product:

Using the properties of scalar triple products and cross products:

\[ \bar{a} \cdot \bar{p} = \frac{\bar{a} \cdot (\bar{b} \times \bar{c})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1 \]

Similarly:

\[ \bar{b} \cdot \bar{q} = 1, \quad \bar{c} \cdot \bar{r} = 1 \]

For the other terms:

\[ \bar{b} \cdot \bar{p} = 0, \quad \bar{c} \cdot \bar{q} = 0, \quad \bar{a} \cdot \bar{r} = 0 \]

because \( \bar{b} \) is perpendicular to \( \bar{b} \times \bar{c} \), etc.

3. Evaluate the expression:

\[ (\bar{a} + \bar{b}) \cdot \bar{p} + (\bar{b} + \bar{c}) \cdot \bar{q} + (\bar{c} + \bar{a}) \cdot \bar{r} = (1 + 0) + (1 + 0) + (1 + 0) = 3 \]

Correct Answer: (D) 3

Answer 2

We are given that:

• $ \mathbf{p} = \frac{\mathbf{a} \times \mathbf{c}}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} $
• $ \mathbf{q} = \frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} $
• $ \mathbf{r} = \frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} $

Substitute these into the expression we need to evaluate:

$$ (\mathbf{a} + \mathbf{b}) \cdot \mathbf{p} + (\mathbf{b} + \mathbf{c}) \cdot \mathbf{q} + (\mathbf{c} + \mathbf{a}) \cdot \mathbf{r} $$ $$ = (\mathbf{a} + \mathbf{b}) \cdot \frac{\mathbf{a} \times \mathbf{c}}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} + (\mathbf{b} + \mathbf{c}) \cdot \frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} + (\mathbf{c} + \mathbf{a}) \cdot \frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} $$

We can factor out $ \frac{1}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} $:

$$ = \frac{1}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} \cdot \left[ (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{c}) + (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{c} \times \mathbf{a}) + (\mathbf{c} + \mathbf{a}) \cdot (\mathbf{a} \times \mathbf{b}) \right] $$

Recall that the scalar triple product is cyclic:

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

And also that:

$$ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = -\mathbf{a} \cdot (\mathbf{c} \times \mathbf{b}) $$

3. Simplifying the expression:

Let's expand the terms within the brackets:

$$ (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{c}) = \mathbf{a} \cdot (\mathbf{a} \times \mathbf{c}) + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{c}) = 0 + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{a} \times \mathbf{c}) $$ $$ (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) + 0 = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) $$ $$ (\mathbf{c} + \mathbf{a}) \cdot (\mathbf{a} \times \mathbf{b}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) + \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) + 0 = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) $$

Substitute these back into the expression:

$$ = \frac{1}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} \cdot \left[ \mathbf{b} \cdot (\mathbf{a} \times \mathbf{c}) + \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) \right] $$

Using the cyclic property:

$$ = \frac{1}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} \cdot \left[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) + \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) + \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \right] $$ $$ = \frac{1}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} \cdot 3[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})] $$ $$ = \frac{1}{[\mathbf{a} \, \mathbf{b} \, \mathbf{c}]} \cdot 3[\mathbf{a} \, \mathbf{b} \, \mathbf{c}] = 3 $$

Therefore, the final result is $ 3 $.