The vector equation of the straight line passing through the point with position vector -3-k- and parallel to the vector- 2- - 3- - 4k- in standard cartesian form is-

The vector equation of the straight line is r=î 3ĵ+k̂+t(2î+3ĵ 4k̂) or xî+yĵ+zk̂ =(1+2t)î+( 3+3t)ĵ+(1 4t)k̂. Eliminating t from each component, we obtain the car ...

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Standard XII

Mathematics

Condition for a Line to Lie on a Plane

The vector eq...

Question

The vector equation of the straight line passing through the point with position vector $^i - 3^j +^k$ and parallel to the vector, $2^i + 3^j - 4^k$ in standard cartesian form is:

\frac{x + 1}{2} = \frac{y + 3}{3} = \frac{z - 1}{- 4}

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\frac{x - 1}{2} = \frac{y + 3}{3} = \frac{z - 1}{4}

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\frac{x - 1}{2} = \frac{y + 3}{3} = \frac{z - 1}{- 4}

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\frac{x - 1}{2} = \frac{y - 3}{3} = \frac{z + 1}{- 4}

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Solution

The correct option is C $\frac{x - 1}{2} = \frac{y + 3}{3} = \frac{z - 1}{- 4}$
The vector equation of the straight line is
$\to r =^i - 3^j +^k + t (2^i + 3^j - 4^k)$
or
$x^i + y^j + z^k = (1 + 2 t)^i + (- 3 + 3 t)^j + (1 - 4 t)^k .$
Eliminating t from each component, we obtain the cartesian form of the straight line,
$\frac{x - 1}{2} = \frac{y + 3}{3} = \frac{z - 1}{- 4} .$

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Similar questions

Q. Find the vector equation of a line passing through a point with position vector

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and parallel to the line joining the points with position vectors

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The vector equation of the straight line passing through the point with position vector ^i−3^j+^k and parallel to the vector, 2^i+3^j−4^k in standard cartesian form is:

The correct option is C x−12=y+33=z−1−4The vector equation of the straight line is →r=^i−3^j+^k+t(2^i+3^j−4^k) or x^i+y^j+z^k=(1+2t)^i+(−3+3t)^j+(1−4t)^k. Eliminating t from each component, we obtain the cartesian form of the straight line, x−12=y+33=z−1−4.

The vector equation of the straight line passing through the point with position vector $^i - 3^j +^k$ and parallel to the vector, $2^i + 3^j - 4^k$ in standard cartesian form is: