Question

$\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar and mutually perpendicular vectors of same magnitude K. $\vec{r}$ is any vector satisfying $\vec{a}\times((\vec{r}-\vec{b})\times\vec{a}) + \vec{b}\times((\vec{r}-\vec{c})\times\vec{b}) + \vec{c}\times((\vec{r}-\vec{a})\times\vec{c}) = \vec{0}$, then $\vec{r} =$

• A. $\frac{K^2(\vec{a}+\vec{b}+\vec{c})}{3K^2-1}$
• B. $\frac{\vec{a}+\vec{b}+\vec{c}}{2}$
• C. $\frac{K(\vec{a}+\vec{b}+\vec{c})}{K+1}$
• D. $\frac{\vec{a}+\vec{b}+\vec{c}}{K^2+1}$

Hint:

The vector triple product identity $\vec{A}\times(\vec{B}\times\vec{C}) = (\vec{A}\cdot\vec{C})\vec{B} - (\vec{A}\cdot\vec{B})\vec{C}$ is crucial for simplifying nested cross products. Also, remember the formula for expressing any vector in terms of an orthogonal basis.

Answer 1

The Correct Option is B

Step 1: Use the vector triple product identity.
The identity is $\vec{A}\times(\vec{B}\times\vec{C}) = (\vec{A}\cdot\vec{C})\vec{B} - (\vec{A}\cdot\vec{B})\vec{C}$.
Let's apply this to the first term of the given equation: $\vec{a}\times((\vec{r}-\vec{b})\times\vec{a})$.
Here, $\vec{A}=\vec{a}, \vec{B}=\vec{r}-\vec{b}, \vec{C}=\vec{a}$.
The term becomes $(\vec{a}\cdot\vec{a})(\vec{r}-\vec{b}) - (\vec{a}\cdot(\vec{r}-\vec{b}))\vec{a}$.

Step 2: Use the given properties of $\vec{a, \vec{b}, \vec{c}$.}
They are mutually perpendicular, so $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0$.
They have the same magnitude K, so $|\vec{a}|^2=|\vec{b}|^2=|\vec{c}|^2 = K^2$. This also means $\vec{a}\cdot\vec{a}=\vec{b}\cdot\vec{b}=\vec{c}\cdot\vec{c}=K^2$.

Step 3: Simplify all three terms of the equation.
First term: $K^2(\vec{r}-\vec{b}) - (\vec{a}\cdot\vec{r} - \vec{a}\cdot\vec{b})\vec{a} = K^2(\vec{r}-\vec{b}) - (\vec{a}\cdot\vec{r})\vec{a}$.
Second term: $K^2(\vec{r}-\vec{c}) - (\vec{b}\cdot\vec{r} - \vec{b}\cdot\vec{c})\vec{b} = K^2(\vec{r}-\vec{c}) - (\vec{b}\cdot\vec{r})\vec{b}$.
Third term: $K^2(\vec{r}-\vec{a}) - (\vec{c}\cdot\vec{r} - \vec{c}\cdot\vec{a})\vec{c} = K^2(\vec{r}-\vec{a}) - (\vec{c}\cdot\vec{r})\vec{c}$.

Step 4: Sum the simplified terms and set to zero.
$K^2(\vec{r}-\vec{b}) - (\vec{a}\cdot\vec{r})\vec{a} + K^2(\vec{r}-\vec{c}) - (\vec{b}\cdot\vec{r})\vec{b} + K^2(\vec{r}-\vec{a}) - (\vec{c}\cdot\vec{r})\vec{c} = \vec{0}$.
$3K^2\vec{r} - K^2(\vec{a}+\vec{b}+\vec{c}) - [(\vec{a}\cdot\vec{r})\vec{a} + (\vec{b}\cdot\vec{r})\vec{b} + (\vec{c}\cdot\vec{r})\vec{c}] = \vec{0}$.

Step 5: Express $\vec{r$ in terms of the basis vectors $\vec{a}, \vec{b}, \vec{c}$.}
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar and mutually perpendicular, they form an orthogonal basis. Any vector $\vec{r}$ can be written as a linear combination. The expression in the square brackets is related to the projection of $\vec{r}$.
The general formula for a vector $\vec{r}$ in terms of an orthogonal basis $\{\vec{v_1}, \vec{v_2}, \vec{v_3}\}$ is $\vec{r} = \frac{\vec{r}\cdot\vec{v_1}}{|\vec{v_1}|^2}\vec{v_1} + \frac{\vec{r}\cdot\vec{v_2}}{|\vec{v_2}|^2}\vec{v_2} + \frac{\vec{r}\cdot\vec{v_3}}{|\vec{v_3}|^2}\vec{v_3}$.
In our case, $\vec{r} = \frac{\vec{r}\cdot\vec{a}}{K^2}\vec{a} + \frac{\vec{r}\cdot\vec{b}}{K^2}\vec{b} + \frac{\vec{r}\cdot\vec{c}}{K^2}\vec{c}$.
This implies $[(\vec{a}\cdot\vec{r})\vec{a} + (\vec{b}\cdot\vec{r})\vec{b} + (\vec{c}\cdot\vec{r})\vec{c}] = K^2\vec{r}$.

Step 6: Substitute this back into the main equation and solve for $\vec{r$.}
$3K^2\vec{r} - K^2(\vec{a}+\vec{b}+\vec{c}) - K^2\vec{r} = \vec{0}$.
$2K^2\vec{r} - K^2(\vec{a}+\vec{b}+\vec{c}) = \vec{0}$.
Since $K \neq 0$, we can divide by $K^2$.
$2\vec{r} - (\vec{a}+\vec{b}+\vec{c}) = \vec{0}$.
$2\vec{r} = \vec{a}+\vec{b}+\vec{c}$.
$\vec{r} = \frac{\vec{a}+\vec{b}+\vec{c}}{2}$.