Let a, b and c are the position vectors of the vertices A, B are C, respectively.

AP EAPCET 2022 - 4th July Evening Shift Mathematics - Three Dimensional Geometry Question 82 English Explanation

Also $$\mathbf{d}, \mathbf{e}$$ and $$\mathbf{f}$$ be the position vectors of points $$D, E$$ and $$F$$, respectively

$$\mathrm{d}=\frac{2 \mathrm{c}+3 \mathrm{~b}}{5}, \mathrm{e}=\frac{2 \mathrm{c}+\mathrm{a}}{3}, \mathrm{f}=\frac{3 \mathrm{~b}+\mathrm{a}}{4}$$

Equation of line $$B E$$ is $$\mathbf{r}=\mathbf{b}+k(\mathbf{e}-\mathbf{b})$$,

$$\begin{aligned} \mathbf{r} & =\mathbf{b}+k\left(\frac{2 \mathbf{c}+\mathbf{a}}{3}-\mathbf{b}\right) \\ \Rightarrow \quad \mathbf{r} & =\mathbf{b}+k\left[\frac{2 \mathbf{c}+\mathbf{a}-3 \mathbf{b}}{3}\right] \\ \Rightarrow \quad \mathbf{r} & =\frac{k}{3} \mathbf{a}+\mathbf{b}[1-k]+\frac{2 k}{3} \mathbf{c} \end{aligned}$$

Equation of line $$C F$$ is

$$\begin{aligned} \mathbf{r} & =\mathbf{c}+\lambda(\mathbf{f}-\mathbf{c}) \\ & =\mathbf{c}+\lambda\left(\frac{3 b+\mathbf{a}}{4}-\mathbf{c}\right)=\mathbf{c}+\lambda\left[\frac{3 b+\mathbf{a}-4 \mathbf{c}}{4}\right] \\ & =\frac{\lambda \mathbf{a}}{4}+\frac{3 \lambda \mathbf{b}}{4}+\mathbf{c}(1-\lambda) \end{aligned}$$

$$C F$$ and $$B E$$ intersect at point $$P$$

$$\frac{k \mathbf{a}}{3}+\mathbf{b}(1-k)+\frac{2 k}{3} \mathbf{c}=\frac{\lambda \mathbf{a}}{4}+\frac{3 \lambda \mathbf{b}}{4}+\mathbf{c}(1-\lambda)$$

On comparing,

$$\frac{k}{3}=\frac{\lambda}{4}$$

$$\Rightarrow \quad k=\frac{3 \lambda}{4}$$

Also,

$$\begin{array}{rlrl} & & 1-k & =\frac{3 \lambda}{4} \Rightarrow 1-k=k \\ \Rightarrow & & 2 k & =1 \Rightarrow k=\frac{1}{2} \\ \text{and} & & \frac{1}{2} & =\frac{3 \lambda}{4} \\ && \lambda & =\frac{2}{3} \end{array}$$

Position vector of $$P=\frac{\mathbf{a}}{6}+\frac{\mathbf{b}}{2}+\frac{\mathbf{c}}{3}=\frac{\mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}$$

$$\begin{aligned} & \mathbf{A P}=x_1 \mathbf{A B}+y_1 \mathbf{A C} \\ &\left(\frac{\mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}-\mathbf{a}\right)=x_1(\mathbf{b}-\mathbf{a})+y_1(\mathbf{c}-\mathbf{a}) \\ & \Rightarrow \frac{-5 \mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}=\mathbf{a}\left(-x_1-y_1\right)+x_1 \mathbf{b}+y_1 \mathbf{C} \\ & \therefore \quad-x_1-y_1=\frac{-5}{6} \\ & x_1+y_1=\frac{5}{6} \end{aligned}$$

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