Question

If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c\hat{k}(a,b,c\ne 1)$ are coplanar, then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to

• A. $0$
• B. $1$
• C. $a+b+c$
• D. $abc$

Answer 1

Answer: (B)

$1$

Three vectors are coplanners means $\overrightarrow{A}=x\overrightarrow{B}+y\overrightarrow{C}a\hat{i}+\hat{j}+\hat{k}=x(\hat{i}+b\hat{j}+\hat{k})+y(\hat{i}+\hat{j}+c\hat{k})$

comparing the coefficient of $\hat{i},\hat{j},\hat{k}$

We get $a=x+yb=\frac{1-y}{x}c=\frac{1-x}{y}$

$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=\frac{1}{1-x-y}+\frac{1}{1-\frac{1-y}{x}}+\frac{1}{1-\frac{1-x}{y}}=\frac{1}{1-x-y}+\frac{x}{x-1+y}+\frac{y}{y-1+x}=\frac{1}{1-x-y}-\frac{x}{1-x-y}-\frac{y}{1-x-y}=\frac{1-x-y}{1-x-y}=1$

So correct answer should be option B