The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2 x + 3 y + 4 z = 5 which is perpendicular to the plane x - y + z = 0 is:

Question

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is:

Smart Achievers · Accepted Answer

(3)

$Let the plane be$ $(x + y + z - 1) + λ (2 x + 3 y + 4 z - 5) = 0$

$\Rightarrow (2 λ + 1) x + (3 λ + 1) y + (4 λ + 1) z - (5 λ + 1) = 0$

$⊥ to the plane$ $x - y + z = 0$

$\Rightarrow λ = - \frac{1}{3}$

$\Rightarrow the required plane is x - z + 2 = 0$

Solution

Q.
The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is:

1 $\vec{r} \times (\hat{i} + \hat{k}) + 2 = 0$

2 $\vec{r} \cdot (\hat{i} - \hat{k}) - 2 = 0$

3 $\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$

4 $\vec{r} \times (\hat{i} - \hat{k}) + 2 = 0$

Ans.
(3)

$Let the plane be$ $(x + y + z - 1) + λ (2 x + 3 y + 4 z - 5) = 0$

$\Rightarrow (2 λ + 1) x + (3 λ + 1) y + (4 λ + 1) z - (5 λ + 1) = 0$

$⊥ to the plane$ $x - y + z = 0$

$\Rightarrow λ = - \frac{1}{3}$

$\Rightarrow the required plane is x - z + 2 = 0$

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Solution

Q. The vector equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0 is:

1 r→×(i^+k^)+2=0

2 r→·(i^-k^)-2=0

3 r→·(i^-k^)+2=0

4 r→×(i^-k^)+2=0

Ans. (3) Let the plane be (x+y+z-1)+λ(2x+3y+4z-5)=0 ⇒(2λ+1)x+(3λ+1)y+(4λ+1)z-(5λ+1)=0 ⊥ to the plane x-y+z=0 ⇒λ=-13 ⇒the required plane is x-z+2=0