MHT CET 2025 26th April Evening Shift | Three Dimensional Geometry Question 1 | Mathematics

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the XY plane and the YZ plane at points A and B respectively. The equation of line through the points A and B is

$[\overline{\mathrm{r}}-(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}})] \times\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{2} \hat{\mathrm{k}}\right)=\overline{0}$

$[\overline{\mathrm{r}}+(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}})] \times\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)=\overline{0}$

$\overline{\mathrm{r}}=(-\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}})+\lambda\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{2} \hat{\mathrm{k}}\right)$

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

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The distance of the point $\mathrm{A}(3,-4,5)$ from the plane $2 x+5 y-6 z=16$ measured along the line $\frac{x}{2}=\frac{y}{1}=\frac{z}{-2}$ is

$\frac{60}{7}$ units

$\frac{7}{60}$ units

$\frac{1}{7}$ units

$\frac{2}{7}$ units

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The direction cosines of a normal to the plane passing through $(4,2,3),(-1,4,2)$ and $(3,2,1)$ are …..

$\quad \frac{-2}{\sqrt{101}}, \frac{3}{\sqrt{101}}, \frac{8}{\sqrt{101}}$

$\frac{-3}{\sqrt{49}}, \frac{2}{\sqrt{49}}, \frac{6}{\sqrt{49}}$

$\quad \frac{-4}{\sqrt{101}}, \frac{-9}{\sqrt{101}}, \frac{2}{\sqrt{101}}$

$\frac{4}{22}, \frac{-12}{22}, \frac{18}{22}$

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