Question

Show that the lines r=(i+2j+3k)+lambda(2i-2j+k) and r=(i+2j+3k)+miu(i+2j+k) are perpendicular to each other.

Answer 1

SOLUTION

TO PROVE

Below two lines are perpendicular to each other

$\displaystyle\sf{ \vec{r} = ( \hat{i} + 2 \hat{j} + 3\hat{k}) + \lambda(2\hat{i} - 2 \hat{j} + \hat{k})}$

$\displaystyle\sf{ \vec{r} = ( \hat{i} + 2 \hat{j} + 3\hat{k}) + \mu(\hat{i} + 2 \hat{j} + 2 \hat{k})}$

EVALUATION

Here the given equation of the lines are

$\displaystyle\sf{ \vec{r} = ( \hat{i} + 2 \hat{j} + 3\hat{k}) + \lambda(2\hat{i} - 2 \hat{j} + \hat{k})}$

$\displaystyle\sf{ \vec{r} = ( \hat{i} + 2 \hat{j} + 3\hat{k}) + \mu(\hat{i} + 2 \hat{j} + 2 \hat{k})}$

So direction ratios of the first line is

$\displaystyle\sf{a_1 = 2 , b_1 = - 2 , c_1 =1 }$

Again direction ratios of the second line is

$\displaystyle\sf{a_2 = 1 , b_2 = 2 , c_2 = 2 }$

Now

$\sf{a_1a_2 + b_1b_2+ c_1c_2}$

$= 2 - 4 + 2$

$= 0$

Hence the given two lines are perpendicular to each other

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