Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ is given by:

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

Rearranging the terms, we get:

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

The normal vector to this plane is $\overrightarrow{n_1}=(1+2\lambda )\hat{i}+(1+3\lambda )\hat{j}+(1+4\lambda )\hat{k}$.

The plane is perpendicular to the plane $x-y+z=0$, which has the normal vector $\overrightarrow{n_2}=\hat{i}-\hat{j}+\hat{k}$.

Since the planes are perpendicular, the dot product of their normal vectors must be zero:

$\overrightarrow{n_1}\cdot \overrightarrow{n_2}=0$

$(1+2\lambda )(1)+(1+3\lambda )(-1)+(1+4\lambda )(1)=0$

$1+2\lambda -1-3\lambda +1+4\lambda =0$

$3\lambda +1=0⟹\lambda =-\frac{1}{3}$

Substituting $\lambda =-\frac{1}{3}$ into the equation of the plane, we get:

$x\left(1-\frac{2}{3}\right)+y\left(1-1\right)+z\left(1-\frac{4}{3}\right)-\left(1-\frac{5}{3}\right)=0$

$\frac{1}{3}x+0y-\frac{1}{3}z-\left(-\frac{2}{3}\right)=0$

Multiplying the entire equation by $3$, we obtain:

$x-z+2=0$

In vector form, this can be written as:

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

Answer: $\overline{\mathrm{r}}\cdot (\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0$