Question

The three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ be such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$, $\vec{a} \cdot \vec{c} = 7$ and $\vec{b}$ is perpendicular to $\vec{c}$, where $\vec{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{k}$, then the value of $2|\vec{a} + \vec{b} + \vec{c}|^2$ is _____.

• A. 75
• B. 150
• C. 37.5
• D. 60

Answer 1

Answer: (A)

75

Answer 2

1. Coplanarity and Perpendicularity: Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$, we can write $\vec{c} = x\vec{a} + y\vec{b}$ for some scalars $x$ and $y$. Given $\vec{b} \cdot \vec{c} = 0$, we substitute $\vec{c}$: $\vec{b} \cdot (x\vec{a} + y\vec{b}) = 0$ $x(\vec{b} \cdot \vec{a}) + y(\vec{b} \cdot \vec{b}) = 0$ Calculate dot products: $\vec{a} = -\hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + \hat{k}$ $\vec{b} \cdot \vec{a} = (2)(-1) + (0)(1) + (1)(1) = -1$ $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 2^2 + 0^2 + 1^2 = 5$ Substituting these: $x(-1) + y(5) = 0 \implies -x + 5y = 0 \implies x = 5y$. So, $\vec{c} = 5y\vec{a} + y\vec{b} = y(5\vec{a} + \vec{b})$.

2. Using $\vec{a} \cdot \vec{c} = 7$: Substitute $\vec{c} = y(5\vec{a} + \vec{b})$ into $\vec{a} \cdot \vec{c} = 7$: $\vec{a} \cdot [y(5\vec{a} + \vec{b})] = 7$ $y [5(\vec{a} \cdot \vec{a}) + (\vec{a} \cdot \vec{b})] = 7$ Calculate required dot products: $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = (-1)^2 + 1^2 + 1^2 = 3$ $\vec{a} \cdot \vec{b} = -1$ (calculated earlier) Substituting these: $y [5(3) + (-1)] = 7 \implies y(15 - 1) = 7 \implies 14y = 7 \implies y = \frac{1}{2}$. Thus, $\vec{c} = \frac{1}{2}(5\vec{a} + \vec{b})$.

3. Calculating $\vec{a} + \vec{b} + \vec{c}$: $\vec{a} + \vec{b} + \vec{c} = \vec{a} + \vec{b} + \frac{1}{2}(5\vec{a} + \vec{b})$ $= (1 + \frac{5}{2})\vec{a} + (1 + \frac{1}{2})\vec{b}$ $= \frac{7}{2}\vec{a} + \frac{3}{2}\vec{b}$

4. Calculating $|\vec{a} + \vec{b} + \vec{c}|^2$: Let $\vec{S} = \frac{7}{2}\vec{a} + \frac{3}{2}\vec{b}$. $|\vec{S}|^2 = (\frac{7}{2}\vec{a} + \frac{3}{2}\vec{b}) \cdot (\frac{7}{2}\vec{a} + \frac{3}{2}\vec{b})$ $= (\frac{7}{2})^2 |\vec{a}|^2 + (\frac{3}{2})^2 |\vec{b}|^2 + 2(\frac{7}{2})(\frac{3}{2})(\vec{a} \cdot \vec{b})$ Using $|\vec{a}|^2 = 3$, $|\vec{b}|^2 = 5$, $\vec{a} \cdot \vec{b} = -1$: $|\vec{S}|^2 = \frac{49}{4}(3) + \frac{9}{4}(5) + 2(\frac{21}{4})(-1)$ $|\vec{S}|^2 = \frac{147}{4} + \frac{45}{4} - \frac{42}{4} = \frac{147 + 45 - 42}{4} = \frac{150}{4} = \frac{75}{2}$.

5. Final Value: The question asks for $2|\vec{a} + \vec{b} + \vec{c}|^2$. $2 \times |\vec{S}|^2 = 2 \times \frac{75}{2} = 75$.