Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vector. Then the point of intersection of the line…

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vector. Then the point of intersection of the line joining the points $\vec{a}+\vec{b}+\vec{c}, \vec{a}-\vec{b}+3 \vec{c}$ and the line joining the points $2 \vec{a}-\vec{b}+\vec{c}, \vec{a}-2 \vec{b}+4 \vec{c}$ is
  1. $2 \vec{a}-4 \vec{c}$
  2. $3 \vec{a}-3 \vec{b}+5 \vec{c}$
  3. $\vec{a}-2 \vec{b}+4 \vec{c}$
  4. $\vec{a}-\vec{b}+3 \vec{c}$

Solution

(c) Let $\left.\begin{aligned} \overrightarrow{O A} & =\vec{a}+\vec{b}+\vec{c} \\ \overrightarrow{O B} & =\vec{a}-\vec{b}+3 \vec{c}\end{aligned} \right\rvert\, \begin{aligned} & \overrightarrow{O C}=2 \vec{a}-\vec{b}+\vec{c} \\ & \overrightarrow{O D}=\vec{a}-2 \vec{b}+4 \vec{c}\end{aligned}$ Hence vector equation of line joining the points $\overrightarrow{O A}$ and $\overrightarrow{O B}$, $ \begin{aligned} & \Rightarrow \quad \vec{r}=\overrightarrow{O A}+\lambda_1(\overrightarrow{O B}-\overrightarrow{O A}), \lambda_1 \in R \\ & \Rightarrow \quad \vec{r}=(\vec{a}+\vec{b}+\vec{c})+\lambda_1(-2 \vec{b}+2 \vec{c}) \end{aligned} $ $\Rightarrow \quad \vec{r}=\vec{a}(1)+\vec{b}\left(1-2 \lambda_1\right)+\vec{c}\left(1+2 \lambda_1\right)$ and also vector equation of line joining the point $\overrightarrow{O C}$ and $\overrightarrow{O D}$ ...(i) $ \begin{aligned} & \vec{r}=\overrightarrow{O C}+\lambda_2(\overrightarrow{O D}-\overrightarrow{O C}), \lambda_2 \in R \\ & \Rightarrow \quad \vec{r}=(2 \vec{a}-\vec{b}+\vec{c})+\lambda_2(-\vec{a}-\vec{b}+3 \vec{c}) \end{aligned} $ $\Rightarrow \quad \vec{r}=\vec{a}\left(2-\lambda_2\right)+\vec{b}\left(-1-\lambda_2\right)+\vec{c}\left(1+\lambda_2\right)$ ...(ii) When the two line will intersect, at that point coefficient of $\vec{a} \cdot \vec{b}$ and $\vec{c}$ will be equal for equation (i) and equation (ii). $ \begin{array}{rl} 2-\lambda_2=1 & 1-2 \lambda_1=-1-\lambda_2 \\ \Rightarrow \lambda_2=1 & \Rightarrow 2 \lambda_1=1+1+\lambda_2=2+1 \\ & \Rightarrow \lambda_1=\frac{3}{2} \end{array} $ Hence the required point of intersection $ \begin{aligned} & \vec{r}=\vec{a}+\vec{b}\left[1-2\left(\frac{3}{2}\right)\right]+\vec{c}\left[1+2\left(\frac{3}{2}\right)\right] \\ & \vec{r}=\vec{a}-2 \vec{b}+4 \vec{c} \end{aligned} $

Asked in: AP EAMCET 2023 (19 May Shift 1)

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