Find the parametric and symmetric equation of the line passing through the point 2-3-4 and perpendicular to the plane 5 x -6 y -7 z -20 -A- x -5 t -2- y -6 t -3- z -7 t -4x-25-y-36-z-47B- x-5 t-2- y-6 t-3- z-7 t-4x-25-y-3-6-x-4-7C- x -5 t -2- y -6 t -3- z -7 t -4x-25-y-3-6-z-4-7D- x -5 t -2- y -6 t -3- z -7 t -4x-25-y-36-z-4-7

The vector perpendicular to the plane, 5x+6y−7z=20 nbsp;is: n→=5i^+6j^ 7j^ This allows us to write the point vector form of the line passing through the poi ...

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Standard XII

Mathematics

Reduction of Non-Symmetrical Form to Symmetrical Form

Find the para...

Question

Find the parametric and symmetric equation of the line passing through the point $(2, 3, 4)$ and perpendicular to the plane $5 x + 6 y - 7 z = 20$ ?

(x = 5 t + 2, y = 6 t + 3, z = - 7 t + 4)

(\frac{x - 2}{5} = \frac{y - 3}{6} = \frac{z - 4}{7})

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(x = 5 t + 2, y = 6 t + 3, z = - 7 t + 4)

(\frac{x - 2}{5} = \frac{y - 3}{- 6} = \frac{x - 4}{- 7})

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(x = 5 t + 2, y = - 6 t + 3, z = - 7 t + 4)

(\frac{x - 2}{5} = \frac{y - 3}{- 6} = \frac{z - 4}{- 7})

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(x = 5 t + 2, y = 6 t + 3, z = - 7 t + 4)

(\frac{x - 2}{5} = \frac{y - 3}{6} = \frac{z - 4}{- 7})

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Solution

The correct option is D $(x = 5 t + 2, y = 6 t + 3, z = - 7 t + 4)$
$(\frac{x - 2}{5} = \frac{y - 3}{6} = \frac{z - 4}{- 7})$
The vector perpendicular to the plane, $5 x + 6 y - 7 z = 20$ is:

$\to n = 5 ˆ i + 6 ˆ j - 7 ˆ j$

This allows us to write the point-vector form of the line passing through the point $(2, 3, 4);$
$\to L = 2^i + 3^j + 4^k + t (5^i + 6^j - 7^j)$
From the point-vector form we can extract the 3 parametric equations by observation:

$x = 5 t + 2 y = 6 t + 3 z = - 7 t + 4$

To find the symmetric form we solve each of the parametric equations for t and then set them equal:

$t = \frac{x - 2}{5} t = \frac{y - 3}{6} t = \frac{z - 4}{- 7}$

Setting them equal gives us the symmetric form:

$\frac{x - 2}{5} = \frac{y - 3}{6} = \frac{z - 4}{- 7}$

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Similar questions

Q. Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)

\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii)

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

(xiv) x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Find the value of $4 x^{2} + y^{2} + 25 z^{2} + 4 x y - 10 y z - 20 z x$ when x = 4, y = 3 and z = 2.

Divide -15x³y²z⁴ +25x²y²z² –20x²yz³ by 5xyz.

Q. Equation of the plane through the point (1, 0, –2) and perpendicular to the planes 2x + y – z – 2 = 0 and x – y –z – 3 = 0 is given by :

Q. The projection of the line

\frac{x - 2}{4} = \frac{y + 2}{8} = \frac{z - 3}{6}

in the plane

P_{1} : 2 x + 3 y - 4 z + 6 = 0

is the line on the intersection of planes

P_{1}

and

P_{2}

. If

P_{1}

is perpendicular to

P_{2}

, then equation of the plane

P_{2}

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Find the parametric and symmetric equation of the line passing through the point (2,3,4) and perpendicular to the plane 5x+6y−7z=20 ?

Find the parametric and symmetric equation of the line passing through the point $(2, 3, 4)$ and perpendicular to the plane $5 x + 6 y - 7 z = 20$ ?