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Question
Find the direction cosines of the line x+22 = 2y−76 = 5−z6. then the vector equation of the line through the point A(1, 2, 3) and parallel to the given line can be expressed as →r=^i+2^j+3^k+λ(2^i+3^j−6^k), where λ is real.
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Solution
Given line is x+22 = 2y−76 = 5−z6
→x+22 = y−7/23 = z−5−6
Direction ratios of the line are 2,3,−6
If a,b,c are the direction ratios of a line then direction cosines are a√a2+b2+c2,b√a2+b2+c2,c√a2+b2+c2
∴ direction cosines are 27,37,−67
And the line passing through (1,2,3) and parallel to the given line is x−12 = y−23 = z−3−6
The vector equation of the line can be written as :
→r=^i+2^j+3^k+λ(2^i+3^j−6^k), where λ is real.
→x+22 = y−7/23 = z−5−6
Direction ratios of the line are 2,3,−6
If a,b,c are the direction ratios of a line then direction cosines are a√a2+b2+c2,b√a2+b2+c2,c√a2+b2+c2
∴ direction cosines are 27,37,−67
And the line passing through (1,2,3) and parallel to the given line is x−12 = y−23 = z−3−6
The vector equation of the line can be written as :
→r=^i+2^j+3^k+λ(2^i+3^j−6^k), where λ is real.
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