The equation of the plane which contains the line $\frac{x}{4} = \frac{y}{2} = \frac{z}{1}$ and is perpendicular to the plane containing the lines $\frac{x - 4}{1} = \frac{y - 5}{4} = \frac{z - 9}{2}$ and $\frac{x + 8}{4} = \frac{y + 6}{2} = \frac{z - 2}{1}$

5 x - 8 y - 4 z = 0

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5 x + 8 y - 4 z = 0

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5 x - 8 y + 4 z = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 x + 8 y + 4 z = 0

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Solution

The correct option is A $5 x - 8 y - 4 z = 0$
Normal vector of the plane containing the lines
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 4 & 2 4 & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 7^j - 14^k$
So the equation of the plane
$∣ ∣ ∣ ∣ \begin{matrix} x & y & z 4 & 2 & 1 0 & 7 & - 14 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 5 x - 8 y - 4 z = 0$

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Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the staight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

Q. The equation of the plane containing the straight line

\frac{x}{2} = \frac{y}{3} = \frac{z}{4}

and perpendicular to the plane containing the straight lines

\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3}

is :

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\frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{z}{5}

and perpendicular to the plane

x - y + z + 2 = 0

Q. The equation of the plane passing through the line of intersection of planes

2 x + y - z = 3

and

5 x - 3 y + 4 z + 9 = 0

and parallel to the line

\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 5}{5}

is ?

Q. The equation of the plane which contains the line

\frac{x}{4} = \frac{y}{2} = \frac{z}{1}

and is perpendicular to the plane containing the lines

\frac{x - 4}{1} = \frac{y - 5}{4} = \frac{z - 9}{2}

and

\frac{x + 8}{4} = \frac{y + 6}{2} = \frac{z - 2}{1}

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The equation of the plane which contains the line x4=y2=z1 and is perpendicular to the plane containing the lines x−41=y−54=z−92 and x+84=y+62=z−21

The correct option is A 5x−8y−4z=0Normal vector of the plane containing the lines =∣∣ ∣ ∣∣^i^j^k142421∣∣ ∣ ∣∣=7^j−14^k So the equation of the plane ∣∣ ∣∣xyz42107−14∣∣ ∣∣=0⇒5x−8y−4z=0

The equation of the plane which contains the line $\frac{x}{4} = \frac{y}{2} = \frac{z}{1}$ and is perpendicular to the plane containing the lines $\frac{x - 4}{1} = \frac{y - 5}{4} = \frac{z - 9}{2}$ and $\frac{x + 8}{4} = \frac{y + 6}{2} = \frac{z - 2}{1}$