MHT CET 2023 9th May Morning Shift | Three Dimensional Geometry Question 124 | Mathematics

A vector $$\overrightarrow{\mathrm{n}}$$ is inclined to $$\mathrm{X}$$-axis at $$45^{\circ}$$, $$\mathrm{Y}$$-axis at $$60^{\circ}$$ and at an acute angle to Z-axis If $$\overrightarrow{\mathrm{n}}$$ is normal to a plane passing through the point $$(-\sqrt{2}, 1,1)$$, then equation of the plane is

$$\sqrt{2} x+y+z=0$$

$$x+\sqrt{2} y+z=1$$

$$-\sqrt{2} x+y+2 z=5$$

$$x+y+\sqrt{2} z=1$$

MHT CET 2023 9th May Morning Shift

MCQ (Single Correct Answer)

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

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

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

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

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

MHT CET 2022 11th August Evening Shift

MCQ (Single Correct Answer)

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The distance between parallel lines

$$\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1}$$ and

$$\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$$ is :

$$\frac{2 \sqrt{5}}{3}$$ units

$$\frac{\sqrt{5}}{3}$$ units

$$\frac{5 \sqrt{5}}{3}$$ units

$$\frac{4 \sqrt{5}}{3}$$ units

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