Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The given points are $A(1,2,3)$ and $B(3,4,5)$ The direction ratios of line segment $AB$ is given by $$\begin{array}{l}\left(x_2-x_1\right),\left(y_2-y_1\right),\left(z_2-z_1\right)\\ (3-1),(4-2),(5-3)=(2,2,2)=(1,1,1)\end{array}$$ Since, the plane bisects $AB$ at right angles, $\mathbf{AB}$ is the normal to the plane. Which is n Therefore, $\mathbf{n}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ Let $c$ be the mid-point of $AB$ $$\begin{array}{rl}(\frac{x_1+x_2}{2},& \frac{y_1+y_2}{2},\frac{z_1+z_2}{2})\\ =& c\left(\frac{1+3}{2},\frac{2+4}{2},\frac{3+5}{2}\right)\\ =& c(2,3,4)\end{array}$$ Let this vector be $\mathbf{a}=2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}}$ Hence, the required equation of vector $$\{\mathbf{r}-(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\mathbf{k})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k})=0$$ We know that, $\mathbf{r}=x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+z\hat{\mathbf{k}}$ $$\begin{array}{l}\Rightarrow \{(x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+z\hat{\mathbf{k}})-(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0\\ \Rightarrow (x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+z\hat{\mathbf{k}})\cdot (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})\\ =(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}})\cdot (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})\\ \Rightarrow x+y+z=2+3+4\\ \Rightarrow x+y+z=9\end{array}$$