Question

Directions: The following question has four choices, out of which one or more are correct.Let $\alpha =a\hat{i}+b\hat{j}+c\hat{k},\beta =b\hat{i}+c\hat{j}+a\hat{k}$ and $y=c\hat{i}+a\hat{j}+b\hat{k}$ be three co-planar vectors with $a\ne b$ and$v=\hat{i}+\hat{j}+\hat{k},$ Then $v$ is perpendicular to

• (A) α
• (B) β
• (C) γ
• (D) None of these

Answer 1

The Correct Option is A, D

Explanation:
$$\begin{array}{r}\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0\\ a^3+b^3+c^3-3abc=0\\ (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\end{array}$$$$\begin{array}{c}\frac{1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)=0\\ \Rightarrow a+b+c=0\\ Ora=b=c\end{array}$$But, $a\ne b,$ therefore,$$a+b+c=0$$Now$$\begin{array}{l}\alpha .v=(a\hat{i}+b\hat{j}+c\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=a+b+c=0\\ \beta \cdot v=(b\hat{i}+c\hat{j}+a\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=b+c+a=0\\ \gamma \cdot v=(c\hat{i}+a\hat{j}+b\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=c+a+b=0\end{array}$$Hence, this is the required solution.

Answer 2

Explanation:
| a b c | = 0
a^3 + b^3 + c^3 - 3abc = 0
(a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca) = 0
⇒ a + b + c = 0
Or a ≠ b, therefore,
a + b + c = 0

Now,
α · v = (a î + b ĵ + c k̂) · (î + ĵ + k̂) = a + b + c = 0
β · v = (b î + c ĵ + a k̂) · (î + ĵ + k̂) = b + c + a = 0
γ · v = (c î + a ĵ + b k̂) · (î + ĵ + k̂) = c + a + b = 0

Hence, this is the required solution.