The shortest distance between the lines r=3{i}+5{j}+7{k}+λ({i}+2{j}+{k})and r=-{i}-{j}-{k}+(7{i}-6{j}+{k}) is

Correct Answer is : 46/55{46}/{5 {5}}46/55

Correct Answer is : 46/55{46}/{5 {5}}46/55

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AP EAMCET Engineering 2013 Solved Paper

Section: Mathematics

Question : 114 of 160

Marks: +1, -0

r=3\hat{i}+5\hat{j}+7\hat{k}+\lambda(\hat{i}+2\hat{j}+\hat{k})

r=-\hat{i}-\hat{j}-\hat{k}+\mu(7\hat{i}-6\hat{j}+\hat{k})

Solution:

The given lines are

r=a_1+\lambda b_1, r=a_2+\mu b_2

where,

a_1=3\hat{i}+5\hat{j}+7\hat{k}, b_1=\hat{i}+2\hat{j}+\hat{k}

a_2=-\hat{i}-\hat{j}-\hat{k}, b_2=7\hat{i}-6\hat{j}+\hat{k}

|b_1 \times b_2| = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 7 & -6 & 1 \end{vmatrix}

\Rightarrow \left| \hat{i}(2+6)-\hat{j}(1-7)+\hat{k}(-6-14) \right|

\Rightarrow \left| 8\hat{i}+6\hat{j}-20\hat{k} \right|

\Rightarrow \sqrt{64+36+400} = \sqrt{500} = 10\sqrt{5}

Now,

[ (a_2-a_1) b_1 b_2 ] = (a_2-a_1) \cdot (b_1 \times b_2)

=(-4\hat{i}-6\hat{j}-8\hat{k}) \cdot (8\hat{i}+6\hat{j}-20\hat{k})

=-32-36+160

=160-68=92

= \frac{ \left[ (a_2-a_1) \cdot (b_1 \times b_2) \right] }{ \left| b_1 \times b_2 \right| }

Shortest distance

= \frac{92}{10\sqrt{5}} = \frac{46}{5\sqrt{5}}

\frac{16}{5\sqrt{5}}

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