Question

If the vector equation to a plane passing through the points $A\left(a\right),B\left(b\right),C\left(c\right)$ is $\overrightarrow{r}=x\overrightarrow{a}+y\overrightarrow{b}+z\overrightarrow{c}$, then $x+y+z=$

• A. $0$
• B. $10$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

Let point $A\left(a\right)=(1,0,0)=$$\overrightarrow{a}=\hat{i}$

$pointB\left(b\right)=(1,1,0)=\overrightarrow{b}=\hat{i}+\hat{j}$

$pointC\left(c\right)=(1,1,0)=\overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}$

so from given eq $\overrightarrow{r}=x\overrightarrow{a}+y\overrightarrow{b}+z\overrightarrow{c}$

$\overrightarrow{r}=x\left(\hat{i}\right)+y(\hat{i}+\hat{j})+z(\hat{i}+\hat{j}+\hat{k})$

$\overrightarrow{r}=(x+y+z)\hat{i}+(y+z)\hat{j}+z\hat{k}$

if we pass plane from $A\left(a\right)$ and compare to its eq only $\hat{i}$ is $1$

$x+y+z=1$.