Question ID: #781
Let $\vec{a}=-\hat{i}+\hat{j}+2\hat{k}$, $\vec{b}=\hat{i}-\hat{j}-3\hat{k}$, $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}=\vec{c}\times\vec{a}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to:
- (1) 4
- (2) -4
- (3) -2
- (4) 2
Solution:
We first simplify vector $\vec{d}$ using the vector triple product expansion:
$$ \vec{d} = (\vec{a}\times\vec{b})\times\vec{a} = -(\vec{a}\times(\vec{a}\times\vec{b})) $$
Alternatively, $\vec{d} = (\vec{a}\cdot\vec{a})\vec{b} – (\vec{a}\cdot\vec{b})\vec{a}$.
Calculate scalar products:
$$ |\vec{a}|^2 = (-1)^2 + (1)^2 + (2)^2 = 1 + 1 + 4 = 6 $$
$$ |\vec{b}|^2 = (1)^2 + (-1)^2 + (-3)^2 = 1 + 1 + 9 = 11 $$
$$ \vec{a}\cdot\vec{b} = (-1)(1) + (1)(-1) + (2)(-3) = -1 – 1 – 6 = -8 $$
Substitute these values into the expression for $\vec{d}$:
$$ \vec{d} = 6\vec{b} – (-8)\vec{a} = 6\vec{b} + 8\vec{a} $$
We need to find $(\vec{a}-\vec{b}) \cdot \vec{d}$:
$$ (\vec{a}-\vec{b}) \cdot (8\vec{a} + 6\vec{b}) $$
Expand using distributive property:
$$ = 8(\vec{a}\cdot\vec{a}) + 6(\vec{a}\cdot\vec{b}) – 8(\vec{b}\cdot\vec{a}) – 6(\vec{b}\cdot\vec{b}) $$
$$ = 8|\vec{a}|^2 – 2(\vec{a}\cdot\vec{b}) – 6|\vec{b}|^2 $$
Substitute the values:
$$ = 8(6) – 2(-8) – 6(11) $$
$$ = 48 + 16 – 66 $$
$$ = 64 – 66 = -2 $$
Ans. (3)
Was this solution helpful?
YesNo