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Find the angle between the following pai...

Find the angle between the following pair of lines :
`vec(r) = hati + hatj - hatk + lambda (hati - 3 hatj + 2 hatk)` and
`vec(r) = 2 hat(i) - hat(j) + mu (3 hat(i) + hatj - 2 hatk)`.

A

`pi+cos^(-1) ((2)/(7))`

B

`cos^(-1) ((3)/(7))`

C

` cos^(-1) ((2)/(7))`

D

`pi - cos^(-1) ((2)/(7))`

Text Solution

AI Generated Solution

The correct Answer is:
To find the angle between the two given lines represented in vector form, we can follow these steps: ### Step 1: Identify the direction vectors of the lines The equations of the lines are given as: 1. \(\vec{r_1} = \hat{i} + \hat{j} - \hat{k} + \lambda (\hat{i} - 3\hat{j} + 2\hat{k})\) 2. \(\vec{r_2} = 2\hat{i} - \hat{j} + \mu (3\hat{i} + \hat{j} - 2\hat{k})\) From these equations, we can extract the direction vectors: - For the first line, the direction vector \(\vec{b_1} = \hat{i} - 3\hat{j} + 2\hat{k}\) - For the second line, the direction vector \(\vec{b_2} = 3\hat{i} + \hat{j} - 2\hat{k}\) ### Step 2: Calculate the dot product of the direction vectors The dot product \(\vec{b_1} \cdot \vec{b_2}\) is calculated as follows: \[ \vec{b_1} \cdot \vec{b_2} = (1)(3) + (-3)(1) + (2)(-2) \] Calculating this gives: \[ = 3 - 3 - 4 = -4 \] ### Step 3: Calculate the magnitudes of the direction vectors Next, we calculate the magnitudes of \(\vec{b_1}\) and \(\vec{b_2}\): \[ |\vec{b_1}| = \sqrt{(1)^2 + (-3)^2 + (2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] \[ |\vec{b_2}| = \sqrt{(3)^2 + (1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \] ### Step 4: Use the dot product to find the cosine of the angle Using the dot product and magnitudes, we can find \(\cos \theta\): \[ \cos \theta = \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}| |\vec{b_2}|} \] Substituting the values: \[ \cos \theta = \frac{-4}{\sqrt{14} \cdot \sqrt{14}} = \frac{-4}{14} = -\frac{2}{7} \] ### Step 5: Calculate the angle \(\theta\) To find the angle \(\theta\), we take the inverse cosine: \[ \theta = \cos^{-1}\left(-\frac{2}{7}\right) \] ### Step 6: Final expression for the angle Using the property of inverse cosine: \[ \theta = \pi - \cos^{-1}\left(\frac{2}{7}\right) \] This gives us the angle between the two lines.
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Find the angle between the following pairs of lines : (i) vec(r) = 2 hati - 5 hatj + hatk + lambda (3 hati + 2 hatj + 6 hatk ) and vec(r) = 7 hati - 6 hatk + mu (hati + 2 hatj + 2 hatk) (ii) vec(r) = 3 hati + hatj - 2 hatk + lambda (hati - hatj - 2 hatk ) and vec(r) = 2 hati - hatj - 56 hatk + mu (3 hati - 5 hatj - 4 hatk) .

Determine whether or not the following pairs of lines intersect : vec(r) = (hati - 2 hatj + 3 hatk) + lambda (- hati + hatj - 2 hatk) and vec(r) = ( hati - hatj - hatk) + mu (hati + 2 hatj - 2 hatk) .

Find the shortest distance between the lines: (i) vec(r) = 6 hat(i) + 2 hat(j) + 2 hatk + lambda (hati - 2hatj + 2 hatk) and vec(r) = - 4 hati - hatk + mu (3 hati - 2 hatj - 2 hatk ) (ii) vec(r) = (4 hat(i) - hat(j)) + lambda (hati + 2hatj - 3 hatk) and vec(r) = (hati - hatj + 2hatk) + mu (2 hati + 4 hatj - 5 hatk ) (iii) vec(r) = (hati + 2 hatj - 4 hatk) + lambda (2 hati + 3 hatj + 6 hatk ) and vec(r) = (3 hati + 3 hatj + 5 hatk) + mu (-2 hati + 3 hatj + 6 hatk )

Find the angles between the following pairs of vectors : vec A= -2hati + 2hatj -hatk and vec B=3hati +6hatj +2hatk .

Find the angles between the following pairs of vectors : vec A= 2hati - 4hatj+ 6hatk and vec B=3hati +hatj +2hatk .

Find the angles between the following pairs of vectors : vec A= hati + hatj + hatk and vec B = -2hati-2hatj-2hatk .

Find the shortest distance betwee the lines : vec(r) = (hati + 2 hatj + hatk ) + lambda ( hati - hatj + hatk) and vec(r) = 2 hati - hatj - hakt + mu (2 hati + hatj + 2 hatk) .

Find the angle between the pair of lines bar r = ( 3 hati + 2 hatj - 4hatk ) + lamda ( hati + 2hatj + 2hatk ) and bar r = ( 5hati - 2 hatk ) + mu ( 3hati + 2 hatj + 6 hatk )

Find the shortest distance between lines: vec(r) = 6 hati + 2 hatj + 2 hatk + lambda ( hati - 2 hatj + 2 hatk) and vec(r) = -4 hati - hatk + mu (3 hati - 2 hatj - 2 hatk) .

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