Consider the line L₁ : x+1/3 = y+2/1 = z+1/2 L₂ : x-2/1 = y+2/2 = z-3/3 The unit vector perpendicular to both L₁ and L₂ is [JEE Adv 2008 P2]

Correct Answer is : −i^−7j^+5k^53{-{i}-{7j}+{5k}}{5{3}}53−i^−7j^+5k^

Correct Answer is : −i^−7j^+5k^53{-{i}-{7j}+{5k}}{5{3}}53−i^−7j^+5k^

Test Index

Vector Algebra

Section: Mathematics

Question : 37 of 38

Marks: +1, -0

Consider the line

L_1

\frac{x+1}{3}

\frac{y+2}{1}

\frac{z+1}{2}

L_2

\frac{x-2}{1}

\frac{y+2}{2}

\frac{z-3}{3}

L_1

L_2

Solution:

We have

L_1

\vec{r}

(-i-2j-k)+\lambda(3i+j+2k)

L_2

\vec{r}

(2i-2j+3k) + \mu(i+2j+3k)

Let

L_1

\vec{r}

\vec{a}_1+\lambda\vec{b}_1

L_2

\vec{r}

\vec{a}_2+\mu\vec{b}_2

Then, the vector perpendicular to both

L_1

and

L_2

is given by

\vec{b}_1 \times \vec{b}_2

\begin{vmatrix} \hat{i} & \hat{b} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix}

-\hat{i}-\hat{7j}+\hat{5k}

Hence, the unit vector is

\frac{-\hat{i}-\hat{7j}+\hat{5k}}{5\sqrt{3}}

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