Solveeit Logo
Login

Question

Question: Show that the lines: \[\overrightarrow r = \left( {3i + 2j - 4k} \right) + \lambda \left( {i + 2j ...

Show that the lines:
r→=(3i+2j−4k)+λ(i+2j+2k)\overrightarrow r = \left( {3i + 2j - 4k} \right) + \lambda \left( {i + 2j + 2k} \right) and r→=(5i−2j)+μ(3i+2j+6k)\overrightarrow r = \left( {5i - 2j} \right) + \mu \left( {3i + 2j + 6k} \right) intersect and hence find the point of intersection.

Explanation

Solution

Here we will first rewrite the given vectors and then we will equate the coefficients of i, j and k in order to find the values of λ\lambda and μ\mu from 2 equations and then put them into the third equation formed and if they satisfy the third equation then the given lines intersect. Then we will put the evaluated values back into the given vectors and hence find the point of intersection.

Complete step-by-step answer:
The given lines are:-
r→=(3i+2j−4k)+λ(i+2j+2k)\overrightarrow r = \left( {3i + 2j - 4k} \right) + \lambda \left( {i + 2j + 2k} \right)
r→=(5i−2j)+μ(3i+2j+6k)\overrightarrow r = \left( {5i - 2j} \right) + \mu \left( {3i + 2j + 6k} \right)
Rearranging the given values of the lines we get:-
\Rightarrow$$$\overrightarrow r = \left( {3 + \lambda } \right)\widehat i + \left( {2 + 2\lambda } \right)\widehat j + \left( { - 4 + 2\lambda } \right)\widehat k$$ and \Rightarrow\overrightarrow r = \left( {5 + 3\mu } \right)\widehat i + \left( { - 2 + 2\mu } \right)\widehat j + \left( {0 + 6\mu } \right)\widehat k$$ Now equating the coefficients of i, j and k we get:- $$3 + \lambda = 5 + 3\mu $$…………………… (1) $$2 + 2\lambda = - 2 + 2\mu $$……………………. (2) $$ - 4 + 2\lambda = 6\mu $$……………………………………….. (3) Now we will find the values of $$\lambda $$ and $$\mu $$ from equations 2 and 3. Hence, solving equations 2 and 3 we get:- Subtracting equation 2 from equation 3 we get:- $\Rightarrow - 4 + 2\lambda - 2 - 2\lambda = 6\mu + 2 - 2\mu Simplifying it we get:- $\Rightarrow$$$4\mu = - 6
\Rightarrow$$$ \Rightarrow \mu = \dfrac{{ - 3}}{2}$$ Putting this value in equation 3 we get:- \Rightarrow - 4 + 2\lambda = 6\left( {\dfrac{{ - 3}}{2}} \right)$$ Solving for the value of $$\lambda $$ we get:- $\Rightarrow2\lambda = - 9 + 4 Simplifying it we get:- $\Rightarrow$$$\lambda = \dfrac{{ - 5}}{2}
Now putting these values in equation 1.
If they satisfy the equation then the given lines intersect each other.
Hence putting in the values we get:-
LHS,
\Rightarrow$$$LHS = 3 - \dfrac{5}{2}$$ Solving it we get:- \RightarrowLHS = \dfrac{1}{2}$$ Now evaluating the value of RHS we get:- $\RightarrowRHS = 5 + 3\left( {\dfrac{{ - 3}}{2}} \right) Solving it we get:- $\Rightarrow$$$RHS = \dfrac{1}{2}
Now since LHS=RHS therefore these values satisfy the equation 1 and hence the given lines intersect each other.
Now we will put the values of λ\lambda and μ\mu in one of the lines to get the point of
intersection.
Hence putting these values in r→=(5+3μ)i^+(−2+2μ)j^+(0+6μ)k^\overrightarrow r = \left( {5 + 3\mu } \right)\widehat i + \left( { - 2 +2\mu } \right)\widehat j + \left( {0 + 6\mu } \right)\widehat k we get:-
r→=(5+3(−32))i^+(−2+2(−32))j^+(6(−32))k^\overrightarrow r = \left( {5 + 3\left( {\dfrac{{ - 3}}{2}} \right)} \right)\widehat i + \left( { - 2 + 2\left({\dfrac{{ - 3}}{2}} \right)} \right)\widehat j + \left( {6\left( {\dfrac{{ - 3}}{2}} \right)} \right)\widehat k
Solving it further we get:-
r→=(12)i^+(−5)j^+(−9)k^\overrightarrow r = \left( {\dfrac{1}{2}} \right)\widehat i + \left( { - 5} \right)\widehat j + \left( { - 9}\right)\widehat k

Hence, (12,−5,−9)\left( {\dfrac{1}{2}, - 5, - 9} \right) is the point of intersection.

Note: Students should note that the point of intersection is the point where two lines cut each other.
Also, students should take care of the calculations to get the correct answer.
Also the values of the values of λ\lambda and μ\mu should satisfy the third equation to prove that two lines intersect each other.