Question

The common point of the two straight lines $$\overrightarrow{r_1} = (i - 2j + 3k) + s(2i + j + k) \quad {and} \quad \overrightarrow{r_2} = (-i + 2j + 7k) + t(i + j + k), \quad t, s \in \mathbb{R}$$ is:

• A. $(11, 8, -3)$
• B. $(-11, -8, -3)$
• C. $(11, -8, 3)$
• D. $(11, -8, -3)$
• E. $(9, 8, -3)$

Hint:

To find the common point of two straight lines, equate their parametric equations and solve for the parameters. Then substitute the values of the parameters into either equation to get the common point.

Answer 1

The Correct Option is B

We are given two parametric equations of straight lines: $$\overrightarrow{r_1} = (i - 2j + 3k) + s(2i + j + k)$$ and $$\overrightarrow{r_2} = (-i + 2j + 7k) + t(i + j + k).$$ To find the common point, we equate the two vectors $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$.
Step 1: Express both vector equations in component form: $$\overrightarrow{r_1} = (1 + 2s)i + (-2 + s)j + (3 + s)k$$ and $$\overrightarrow{r_2} = (-1 + t)i + (2 + t)j + (7 + t)k.$$ Step 2: Set the components of $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$ equal to each other: $$1 + 2s = -1 + t, \quad -2 + s = 2 + t, \quad 3 + s = 7 + t.$$ Step 3: Solve the system of equations: From $1 + 2s = -1 + t$, we get: $$t = 2s + 2. \quad {(Equation 1)}$$ From $-2 + s = 2 + t$, we substitute $t = 2s + 2$ into this equation: $$-2 + s = 2 + (2s + 2),$$ $$-2 + s = 2 + 2s + 2 \quad \Rightarrow \quad -2 + s = 4 + 2s,$$ $$s = -6.$$ Substitute $s = -6$ into Equation 1: $$t = 2(-6) + 2 = -12 + 2 = -10.$$ Step 4: Substitute $s = -6$ and $t = -10$ into the vector equations to find the common point. Using $s = -6$ in $\overrightarrow{r_1}$: $$\overrightarrow{r_1} = (1 + 2(-6))i + (-2 + (-6))j + (3 + (-6))k = (-11)i + (-8)j + (-3)k.$$
Thus, the common point is $(-11, -8, -3)$.
Therefore, the correct answer is option (B).