Question

Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be three vectors such that $[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]=4$. If $\overrightarrow{x}=p(\overrightarrow{b}\times\overrightarrow{c})+q(\overrightarrow{c}\times\overrightarrow{a})+r(\overrightarrow{a}\times\overrightarrow{b})$ be perpendicular to $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$, then the value of $\frac{(p-q)^2+(q-r)^2+(r-p)^2}{p^2+q^2+r^2}$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

Given three vectors $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$ with $[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}] = 4$. A vector $\overrightarrow{x}$ is defined as $\overrightarrow{x}=p(\overrightarrow{b}\times\overrightarrow{c})+q(\overrightarrow{c}\times\overrightarrow{a})+r(\overrightarrow{a}\times\overrightarrow{b})$. The condition that $\overrightarrow{x}$ is perpendicular to $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ implies their dot product is zero: $\overrightarrow{x} \cdot (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) = 0$ Substituting the expression for $\overrightarrow{x}$: $(p(\overrightarrow{b}\times\overrightarrow{c})+q(\overrightarrow{c}\times\overrightarrow{a})+r(\overrightarrow{a}\times\overrightarrow{b})) \cdot (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) = 0$ Expanding the dot product, and using the properties of the scalar triple product, specifically $(\vec{u} \times \vec{v}) \cdot \vec{w} = [\vec{u}\vec{v}\vec{w}]$ and that $[\vec{u}\vec{v}\vec{w}] = 0$ if any two vectors are identical, the equation simplifies to: $p[\overrightarrow{b}\overrightarrow{c}\overrightarrow{a}] + q[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}] + r[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}] = 0$ Using the cyclic property of the scalar triple product, $[\overrightarrow{b}\overrightarrow{c}\overrightarrow{a}] = [\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}] = [\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]$. Let $K = [\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}] = 4$. The equation becomes: $pK + qK + rK = 0$ $K(p+q+r) = 0$ Since $K=4 \neq 0$, we must have $p+q+r = 0$.

Now, we evaluate the expression $\frac{(p-q)^2+(q-r)^2+(r-p)^2}{p^2+q^2+r^2}$. Expand the numerator: $(p-q)^2+(q-r)^2+(r-p)^2 = (p^2-2pq+q^2) + (q^2-2qr+r^2) + (r^2-2rp+p^2) \\ = 2(p^2+q^2+r^2) - 2(pq+qr+rp)$ From the condition $p+q+r=0$, squaring both sides gives: $(p+q+r)^2 = 0^2$ $p^2+q^2+r^2 + 2(pq+qr+rp) = 0$ $2(pq+qr+rp) = -(p^2+q^2+r^2)$ Substitute this into the numerator: $\text{Numerator} = 2(p^2+q^2+r^2) - (-(p^2+q^2+r^2)) \\ = 2(p^2+q^2+r^2) + (p^2+q^2+r^2) \\ = 3(p^2+q^2+r^2)$ The expression becomes: $\frac{3(p^2+q^2+r^2)}{p^2+q^2+r^2}$ Assuming $p, q, r$ are not all zero (which would make $\overrightarrow{x}=\overrightarrow{0}$ and the expression $\frac{0}{0}$), $p^2+q^2+r^2 \neq 0$. Therefore, we can cancel the term $p^2+q^2+r^2$: $\text{Value} = 3$