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# Angles Between two Lines in 3D Space

• Difficulty Level : Easy
• Last Updated : 21 Feb, 2021

Straight Lines in 3D space are generally represented in two forms Cartesian Form and Vector Form. Hence the angles between any two straight lines in 3D space are also defined in terms of both the forms of the straight lines. Let’s discuss the methods of finding the angle between two straight lines in both forms one by one.

## Cartesian Form

L1: (x – x1) / a1 = (y – y1) / b1 = (z – z1) / c1

L2: (x – x2) / a2 = (y – y2) / b2 = (z – z2) / c2

Here L1 & L2 represent the two straight lines passing through the points (x1, y1, z1) and (x2, y2, z2) respectively in 3D space in Cartesian Form.

• Direction ratios of line L1 are a1, b1, c1 then a vector parallel to L1 is 1 = a1 i + b1  j + c1 k
• Direction ratios of line L2 are a2, b2, c2 then a vector parallel to L2 is 2 = a2 i + b2  j + c2 k

Then the angle âˆ… between L1 and L2 is given by:

âˆ… = cos-1{(12) / (|1| Ã— |2|)}

### Examples

Example 1: (x – 1) / 1 = (2y + 3) / 3 = (z + 5) / 2 and (x – 2) / 3 = (y + 1) / -2 = (z – 2) / 0 are the two lines in 3D space then the angle âˆ… between them is given by:

Solution:

1 = 1 i + (3 / 2)  j + 2 k

2 = 3 i – 2 j + 0 k

|1| = âˆš(12 + (3/2)2 + 22) = âˆš(29 / 2)

|2| = âˆš(32 + 22 + 02) = âˆš(13)

âˆ…  = cos-1{(1Ã—3 + (3/2)Ã—(-2) + (2)Ã—0 ) / ((âˆš(29) / 2) Ã— âˆš(13))}

âˆ…  = cos-1{0 / ((âˆš(29) / 2) Ã— âˆš(13))}

âˆ…  = cos-1(0)

âˆ…  = Ï€ / 2

Example 2: Find the angles between the two lines in 3D space whose only direction ratios are given 2, 1, 2 and 2, 3, 1. In the question, equations of the 2 lines are not given, only their DRs are given. So the angle âˆ… between the 2 lines is given by:

Solution:

1 = Vector parallel to the line having DRs 2, 1, 2 = (2 i + j + 2 k)

|1| = âˆš(22 + 12 + 22) = âˆš9 = 3

2 = Vector parallel to the line having DRs 2, 3, 1 = (2 i + 3 j + k)

|2| = âˆš(22 + 32 + 12) = âˆš(14)

âˆ…  = cos-1{(2Ã—2 + 1Ã—3 + 2Ã—1) / (3 Ã— âˆš(14))}

âˆ…  = cos-1{(4 + 3 + 2) / (3 Ã— âˆš(14))}

âˆ…  = cos-1{9 / (3 Ã— âˆš(14))}

âˆ…  = cos-1(3 / âˆš(14))

Example 3: (x – 1) / 2 = (y – 2) / 1 = (z – 3) / 2 and (x – 2) / 2 = (y – 1) / 2 = (z – 3) / 1 are the two lines in 3D space then the angle âˆ… between them is given by:

Solution:

1 = 2 i + j + 2 k

|1| = âˆš(22 + 12 + 22) = âˆš9 = 3

2 = 2 i + 2 j + k

|2| = âˆš(22 + 22 + 12) = âˆš9 = 3

âˆ…  = cos-1{(2Ã—2 + 1Ã—2 + 2Ã—1 ) / (3 Ã— 3)}

âˆ…  = cos-1{(4 + 2 + 2) / 9}

âˆ…  = cos-1(8 / 9)

## Vector Form

L1 = 1 + t . 1

L2 = 2 + u . 2

Here L1 & L2 represent the two straight lines passing through the points whose position vectors are 1 and 2 respectively in 3D space in Vector Form. 1 & 2 are the two vectors parallel to L1 and L2 respectively and t & u are the parameters. Then the angle âˆ… between the vectors 1 and 2 is equals to the angle between L1 and L2 is given by:

âˆ… = cos-1{(12) / (|1| Ã— |2|)}

### Examples

Example 1:  = (i + j + k) + t Ã— {(-âˆš3 – 1) i + (âˆš3 – 1) j + 4 k} and  = (i + j + k) + u Ã— (i +  j + 2 k)  are the two lines in 3D space then the angle âˆ… between them is given by:

Solution:

1 = (-âˆš3 – 1) i + (âˆš3 – 1) j + 4 k

|1| = âˆš{(-âˆš3 – 1)2 + (âˆš3 – 1)2 + 42)} = âˆš(24)

2 = i +  j + 2 k

|2| = âˆš(12 + 12 + 22) = âˆš6

âˆ…  = cos-1{(-âˆš3 – 1)Ã—1 + (âˆš3 – 1)Ã—1 + 4Ã—2 ) / (âˆš(24) Ã— âˆš6)}

âˆ…  = cos-1{6 / (âˆš(24) Ã— âˆš6)}

âˆ…  = cos-1(Â½)

âˆ…  = Ï€ / 3

Example 2: (i + 2 j + 2 k) and (3 i + 2 j + 6 k) are the two vectors parallel to the two lines in 3D space then the angle âˆ… between them is given by:

Solution:

1 = i + 2 j + 2 k

|1| = âˆš(12 + 22 + 22)} = âˆš9 = 3

2 = 3 i + 2 j + 6 k

|2| = âˆš(32 + 22 + 62) = âˆš(49) = 7

âˆ…  = cos-1{(1Ã—3 + 2Ã—2 + 2Ã—6) / (7 Ã— 3)}

âˆ…  = cos-1{(3 + 4 + 12) / 21}

âˆ…  = cos-1(19 / 21)

Example 3:  = (3 i + 5 j + 7 k) + s Ã— {(i + 2 j – 2 k} and  = (4 i + 3 j + k) + t Ã— (2 i + 4 j – 4 k)  are the two lines in 3D space then the angle âˆ… between them is given by:

Solution:

1 = i + 2 j – 2 k

|1| = âˆš(12 + 22 + (-2)2)} = âˆš9 = 3

2 = 2 i + 4 j – 4 k

|2| = âˆš(22 + 42 + (-4)2) = âˆš(36) = 6

âˆ…  = cos-1{(1Ã—2 + 2Ã—4 + (-2)Ã—(-4)) / (3 Ã— 6)}

âˆ…  = cos-1{(2 + 8 + 8) / 18}

âˆ…  = cos-1(18 / 18)

âˆ…  = cos-1(1) = 0

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