MHT CET 2024 4th May Evening Shift | Three Dimensional Geometry Question 49 | Mathematics

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

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

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

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

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

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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 $\mathrm{k}=3$

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

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Let $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\mathrm{L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ be two given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

$\frac{-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}}{\sqrt{99}}$

$\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$

$\frac{-\hat{i}+7 \hat{j}+5 \hat{k}}{5 \sqrt{3}}$

$\frac{7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}}{\sqrt{99}}$

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

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Let $a, b \in R$. If the mirror image of the point $\mathrm{p}(\mathrm{a}, 6,9)$ w.r.t. line $\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20, b,-a-9)$, then $|a+b|$ is equal to

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