MHT CET 2021 22th September Evening Shift | Three Dimensional Geometry Question 154 | Mathematics

A line drawn from a point $$A(-2,-2,3)$$ and parallel to the line $$\frac{x}{-2}=\frac{y}{2}=\frac{z}{-1}$$ meets the $$\mathrm{YOZ}$$ plane in point $$\mathrm{P}$$, then the co-ordinates of the point $$\mathrm{P}$$ are

$$(0,4,-4)$$

$$(0,2,2)$$

$$(0,-2,2)$$

$$(0,-4,4)$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

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

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

$$\bar{r}=(-\hat{i}+2 \hat{j}-\hat{k})+\lambda(3 \hat{i}+6 \hat{j}-\hat{k})$$

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

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

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

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The plane $$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$$ cuts the $$X$$-axis at A, Y-axis at B and Z-axis at C, then the area of $$\triangle \mathrm{ABC}=$$

$$\sqrt{71}$$ sq. units

$$\sqrt{29}$$ sq. units

$$\sqrt{41}$$ sq. units

$$\sqrt{61}$$ sq. units

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

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If a plane meets the axes $$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$$ in $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ respectively such that centroid of $$\triangle \mathrm{ABC}$$ is $$(1,2,3)$$, then the equation of the plane is

$$x+2 y+3 z=1$$

$$x+\frac{y}{2}+\frac{z}{3}=3$$

$$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$$

$$\frac{x}{4}+\frac{y}{8}+\frac{z}{12}=1$$

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