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  • Find the angle between the lines vec{r}=3 hat{i}-2hat{j}+6hat{k}+lambda left ( 2 hat{i}+\hat {j}+2hat{k} \right ) and vec{r}= (2 hat{i}-5hat{j} )+mu ( 6har{i}+3hat {j}+2hat{k} )

Find the angle between the lines \vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda \left ( 2 \hat{i}+\hat {j}+2\hat{k} \right ) and \vec{r}=\left (2 \hat{i}-5\hat{k} \right )+\mu \left ( 6\har{i}+3\hat {j}+2\hat{k} \right )

Answers (1)

Given, lines:

\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )

\vec{r}=\left (2\hat{i}-5\hat{k} \right )+\mu\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )

We are instructed to find the angle between the lines.

The line \vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right ) is parallel to the vector

2\hat{i}+\hat{j}+2\hat{k}

Let

\vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}

Then, we can say the line \vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )   is parallel to vector \vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}

Similarly, let \vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}

Then, we can say is \vec{r}=2\hat{j}-5\hat{k}+\mu \left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )  parallel to the vector \vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}

If we take θ as the angle between the lines, then cosine θ is:

\cos \theta = \frac{\vec{b_{1}}\vec{b_{2}}}{\left |\vec{b_{1}} \right |\left |\vec{b_{2}} \right |}

Substituting the values of \vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}  and \vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k} in the above equation,

We get 

\cos \theta=\frac{\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )}{\left | 2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |}

Here,

\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=\left ( 2 \times 6 \right )+\left ( 1 \times 3 \right )+\left ( 2 \times 2 \right )

\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=12+3+4

\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=19...........(i)

Also,

\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{2^{2}+1^{2}+2^{2}}\sqrt{6^{2}+3^{2}+2^{2}}

\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{4+1+4}\sqrt{36+9+4}

\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{9}\sqrt{49}

\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=3 \times 7

\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=21............(ii)

Substituting the values of \cos \theta in equation (i) and (ii), we get

\cos \theta=\frac{19}{21}

\Rightarrow \theta=\cos^{-1}\left (\frac{19}{21} \right )

Therefore, the angle between the lines is \cos^{-1}\left (\frac{19}{21} \right ) 

 

Posted by

infoexpert24

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