Geometry PYQ — Page 4

Consider the line $\frac{x-2}{2} = \frac{2y-5}{-3}, z = -1$. Then which of the following is/are true? (A) It has Direction ratios (2, -3, -1) (B) It has Direction cosines $\left(\frac{4}{5}, \frac{-3}{5}, \frac{-1}{5}\right)$ (C) It has Direction ratios $\left(2, \frac{-3}{2}, 0\right)$ (D) It has Direction cosines $\left(\frac{4}{5}, \frac{-3}{5}, 0\right)$ Choose the correct answer from the options given below:

for $|x| < 1$, $sin (tan^{-1}x)$ equal to

Cosine of the acute angle between the lines $\frac{x-3}{2} = \frac{y-2}{1} = \frac{z-5}{2}$ and $\frac{x-1}{6} = \frac{y-3}{-3} = \frac{z+6}{2}$ is

Match List-I with List-II | List-I | List-II | |---|---| | (A) $\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}$ | (I) $\frac{3\pi}{4}$ | | (B) $\tan^{-1}2 + \tan^{-1}3$ | (II) $\pi$ | | (C) $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3$ | (III) $\tan^{-1}\frac{1}{2}$ | | (D) $\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{13}$ | (IV) $\tan^{-1}\frac{2}{9}$ | Choose the correct answer from the options given below:

Match List-I with List-II | List-I | List-II | |---|---| | (A) Line : $x = 2y + 1 = z - 1$ | (I) Crosses $xz$ plane at (1, 0, 1) | | (B) Line : $x + 1 = 2y + 1 = z$ | (II) Crosses $xz$ plane at (0, 0, 1) | | (C) Line : $x - 1 = 2y = z + 1$ | (III) Crosses $xz$ plane at (1, 0, -1) | | (D) Line : $x - 1 = 2y = z - 1$ | (IV) Crosses $xz$ plane at (1, 0, 2) | Choose the correct answer from the options given below:

The vector equation of line passing through (2, -1, 3) and perpendicular to the lines $\frac{x-2}{3} = \frac{y-1}{1} = \frac{z+2}{2}$ and $\frac{x+3}{-4} = \frac{y-5}{-3} = \frac{z+1}{2}$ is (Here $\lambda$ is a parameter)

If a line makes angles $\alpha$, $\beta$ and $\gamma$ with the positive directions of x-axis, y-axis and z-axis respectively, then $\sin^2\alpha + \sin^2\beta + \sin^2\gamma$ is equal to

The direction ratios of the line perpendicular to the lines $\dfrac{x-5}{2} = \dfrac{y+11}{-3} = \dfrac{z+3}{1}$and $\dfrac{x-7}{1} = \dfrac{y+2}{2} = \dfrac{z-4}{-2}$ are proportional to:

The vector equation of line passing through $(-1, 3, -2)$ and perpendicular to the lines $\frac{x+4}{1} = \frac{y}{2} = \frac{z-3}{3}$ and $\frac{x+2}{-3} = \frac{y+5}{2} = \frac{z-6}{5}$ is

If $\tan ^{-1}\left(\frac{2}{3^{-x}+1}\right)=\cot ^{-1}\left(\frac{3}{3^{x}+1}\right)$, then which one of the following is true ?

The distance between the lines $\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}-2 \hat{j}+1 \hat{k}+\mu(4 \hat{i}+6 \hat{j}+12 \hat{k})$ is :

The direction cosines of the line which is perpendicular to the lines with direction ratios $1,-2,-2$ and $0,2,1$ are :

If $\sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3}$, then the value of $\cos^{-1} x + \cos^{-1} y$ is :

The angle between the lines $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{j} - 2\hat{k} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$ is :

The principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

The value of $\lambda$, so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular is:

Let $\tan^{-1}y = \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. Then $y$ is :

The angle between the line $\frac{x+2}{3} = \frac{y-3}{2} = \frac{z+5}{6}$ and the plane $2x + 10y - 11z = 5$ is:

Match List I with List II | LIST I | LIST II | |---|---| | A. $lx + my + nz = d$ is | I. Equation of plane passing through a given point and normal to given vector | | B. $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ is | II. Equation of plane in normal form | | C. $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ | III. Plane passing through the intersection of two planes | | D. $(a_1x + b_1y + c_1z + d_1) + \lambda(a_2x + b_2y + c_2z + d_2) = 0$ | IV. Intercept from of plane | Choose the correct answer from the options given below:

The vector equation of the line joining the points $(-2, -3, -4)$ and $(1, -2, 4)$ is :

The angle between the two planes $x + y - z = 3$ and $3x + 2y + z = 5$ is :

Cartesian equation of plane passing through the points (2, -4, 5) and perpendicular to the line with direction ratios (3, -1, 2) is :

The shortest distance between the lines $\frac{x+3}{1} = \frac{y-2}{2} = \frac{z+4}{3}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$ is :

The principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is :