मराठी

The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.

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प्रश्न

The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.

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उत्तर

The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is `(x - 3)hat"i" + (y - 4)hat"j" + (z + 7)hat"k" = lambda(-2hat"i" - 5hat"j" + 13hat"k")`.

Explanation:

Given the points (3, 4, – 7) and (1, – 1, 6)

Here `vec"a" = 3hat"i" + 4hat"j" - 7hat"k"` and `vec"b" = hat"i" - hat"j" + 6hat"k"`

Equation of the line is `vec"r" = vec"a" + lambda(vec"b" - vec"a")`

⇒ `vec"r" = (3hat"i" + 4hat"j" - 7hat"k") + lambda[(hat"i" - hat"j" + 6hat"k") - (3hat"i" + 4hat"j" - 7hat"k")]`

⇒ `vec"r" = (3hat"i" + 4hat"j" - 7hat"k") + lambda(-2hat"i" - 5hat"j" + 13hat"k")`

⇒ `(xhat"i" + yhat"j" + zhat"k") = (3hat"i" + 4hat"j" - 7hat"k") + lambda(-2hat"i" - 5hat"j" + 13hat"k")`

⇒ `(x - 3)hat"i" + (y - 4)hat"j" + (z + 7)hat"k" = lambda(-2hat"i" - 5hat"j" + 13hat"k")`

Hence, the vector equation of the line is `(x - 3)hat"i" + (y - 4)hat"j" + (z + 7)hat"k" = lambda(-2hat"i" - 5hat"j" + 13hat"k")`

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पाठ 11: Three Dimensional Geometry - Exercise [पृष्ठ २३९]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 11 Three Dimensional Geometry
Exercise | Q 40 | पृष्ठ २३९
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