The two lines L₁ : {r} = ({i}+{5j}+{5k})+t({4i}-{4j}+{5k}) and L₂ : {r} = ({2i}+{4j}+{5k})+s({8i}-{3j}+{k}) are such that

Correct Answer is : both are Skew lines

both are non-Skew lines, non-parallel, non-perpendicular

Correct Answer is : both are Skew lines

Test Index

TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 32 of 160

Marks: +1, -0

The two lines

L_1

\overline{r}

(\overline{i}+\overline{5j}+\overline{5k})+t(\overline{4i}-\overline{4j}+\overline{5k})

L_2

\overline{r}

(\overline{2i}+\overline{4j}+\overline{5k})+s(\overline{8i}-\overline{3j}+\overline{k})

are such that

Solution: 👈: Video Solution

\; \text{ } \; r=(\hat{i}+5\hat{j}+5\hat{k})+t(4\hat{i}-4\hat{j}+5\hat{k})

\; r=(2\hat{i}+4\hat{j}+5\hat{k})+s(8\hat{i}-3\hat{j}+\hat{k})

Let's find the shortest distance between them projection of

(2-1)\hat{i}+(4-5)\hat{j}+(5-5)\hat{k}

\; \frac{n_{1} \times n_{2}}{|n_{1} \times n_{2}|}

\;\;\; n_{1} \times n_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 5 \\ 8 & -3 & 1 \end{vmatrix} = 11\hat{i}+36\hat{j}+20\hat{k}

\; \text{Perpendicular distance} \; = (\hat{i} - \hat{j}) \cdot \; \frac{11\hat{i}+36\hat{j}+20\hat{k}}{\sqrt{11^{2}+36^{2}+20^{2}}}

Clearly, this is not zero.

\therefore

Both the lines are skew lines.

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