MHT CET 2024 4th May Evening Shift | Three Dimensional Geometry Question 47 | Mathematics

The equation of the plane, passing through the point $(-1,2,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$, is

$8 x-14 y-13 z-3=0$

$8 x-14 y+13 z+75=0$

$8 x+14 y+13 z+19=0$

$8 x+14 y-13 z-59=0$

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

-0

The co-ordinates of the point where the line through $\mathrm{A}(3,4,1)$ and $\mathrm{B}(5,1,6)$ crosses the $x y$-plane are

$\left(\frac{13}{5}, \frac{23}{5}, 0\right)$

$\left(-\frac{13}{5}, \frac{23}{5}, 0\right)$

$\left(\frac{13}{5},-\frac{23}{5}, 0\right)$

$\left(-\frac{13}{5},-\frac{23}{5}, 0\right)$

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

-0

The Cartesian equation of a line is $2 x-2=3 y+1=6 z-2$, then the vector equation of the line is

$\overline{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{\hat{\mathrm{j}}}{3}+\frac{\hat{\mathrm{k}}}{3}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

$\overline{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{3}\right)+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)$

$\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

$\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)$

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

-0

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k} \quad$ and $\frac{x-1}{\mathrm{k}}=\frac{y-4}{2}=\frac{\mathrm{z}-5}{1}$ are coplanar if

$\mathrm{k}=1$ or $\mathrm{k}=-1$

$\mathrm{k}=0$ or $\mathrm{k}=-3$

$\mathrm{k}=3$ or $\mathrm{k}=-3$

$\mathrm{k}=0$ or $\mathrm{k}=3$

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