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If one end of a focal chord AB of the parabola $y^{2}=8x$ is at $A\left ( \frac{1}{2},-2 \right ),$ then the equation of the tangent to it at B is :
Option: 1 $x+2y+8=0$
Option: 2 $2x-y-24=0$
Option: 3 $x-2y+8=0$
Option: 4 $2x+y-24=0$

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Shabareesh

The length of the minor axis (along y-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}.$ If this ellipse touches the line, $x+6y=8;$ then its eccentricity is :
Option: 1 $\frac{1}{2}\sqrt{\frac{5}{3}}$

Option: 2 $\frac{1}{2}\sqrt{\frac{11}{3}}$

Option: 3 $\sqrt{\frac{5}{6}}$

Option: 4 $\frac{1}{3}\sqrt{\frac{11}{3}}$

What is Ellipse? -

Ellipse

Standard Equation of Ellipse:

The standard form of the equation of an ellipse with center (0, 0) and major axis on the x-axis is

$\mathbf{\frac{\mathbf{x}^{2}}{\mathbf{a}^{2}}+\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}=1} \quad \text { where }, \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)_{(\mathrm{a}>\mathrm{b})}$

a > b
the length of the major axis is 2a
the coordinates of the vertices are (±a, 0)
the length of the minor axis is 2b
the coordinates of the co-vertices are (0, ±b)

Equation of Tangent of Ellipse in Parametric Form and Slope Form -

Slope Form:

$\\ {\text { The equation of tangent of slope m to the ellipse, } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { are }} \\ {y=m x \pm \sqrt{a^{2} m^{2}+b^{2}} \text { and coordinate of point of contact is }} \\ {\left(\mp \frac{a^{2} m}{\sqrt{a^{2} m^{2}+b^{2}}}, \pm \frac{b^{2}}{\sqrt{a^{2} m^{2}+b^{2}}}\right)}$

$\\\begin{array}{l}{2 \mathrm{b}=\frac{4}{\sqrt{3}} \quad \Rightarrow \quad \mathrm{b}=\frac{2}{\sqrt{3}}} \\ {\text { Equation of tangent } \equiv \mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}}\end{array}\\\text { comparing with } \equiv y=\frac{-x}{6}+\frac{4}{3}\\$

$\\ {\mathrm{m}=\frac{-1}{6} \text { and } \mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}+\frac{4}{3}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}=\frac{16}{9}-\frac{4}{3}=\frac{4}{9}} \\ {\Rightarrow \quad a^{2}=16} \\ {\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}}\\e=\sqrt{\frac{11}{12}}$

Correct Option (2)

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avinash.dongre

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If $e_{1}\: \: and\: \: e_{2}$ are the eccentricities of the ellipse, $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$ and the hyperbola, $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ respectively and $\left ( e_{1},e_{2} \right )$ is a point on the ellipse, $15x^{2}+3y^{2}=k,$ then $k$ is equal to :
Option: 1 $14$
Option: 2 $15$
Option: 3 $17$
Option: 4 $16$

What is Ellipse? -

Ellipse

$\mathbf{\frac{\mathbf{x}^{2}}{\mathbf{a}^{2}}+\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}=1} \quad \text { where }, \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)_{(\mathrm{a}>\mathrm{b})}$

a > b
the length of the major axis is 2a
the coordinates of the vertices are (±a, 0)
the length of the minor axis is 2b
the coordinates of the co-vertices are (0, ±b)

What is Hyperbola? -

Hyperbola:

Eccentricity of Hyperbola:

$\\\mathrm{Equation\;of\;the\;hyperbola\;is\;\;\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}\\\text{we have,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;b^2=a^2\left ( e^2-1 \right )}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e^2=\frac{b^2+a^2}{a^2}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{b^2}{a^2} \right )}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{2b}{2a} \right )^2}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{conjugate \;axis}{transverse \;axis} \right )^2}}$

$\begin{aligned} &e_{1}=\sqrt{1-\frac{4}{18}}=\sqrt{\frac{7}{9}}=\frac{\sqrt{7}}{3}\\ &\mathrm{e}_{2}=\sqrt{1+\frac{4}{9}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}\\ &15 e_{1}^{2}+3 e_{2}^{2}=k \Rightarrow \quad k=15\left(\frac{7}{9}\right)+3\left(\frac{13}{9}\right) \end{aligned}$

So, k= 16

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avinash.dongre

Let C the centroid of the triangle with vertices $(3,-1),(1,3)\: and \: (2,4).$ Let P be the point of intersection of the lines $x+3y-1 =0$ and $3x-y+1 =0.$ Then the line passing through the points C and P also passes through the point:
Option: 1 $(-9,-7)$
Option: 2 $(-9,-6)$
Option: 3 $(7,6)$
Option: 4 $(9,7)$

Centroid -

Centroid

Centroid of a triangle is the point of intersection of the medians of the triangle. A centroid divides the median in the ratio 2:1.

Whereas, the median is the line joining the mid-points of the sides and the opposite vertices.

The coordinates of the centroid of a triangle (G) whose vertices are A (x₁, y₁), B (x₂, y₂) and C(x₃, y₃), is given by

$\\\mathrm{\mathbf{\left ( \frac{x_1+x_2+x_3}{3},\;\frac{y_1+y_2+y_3}{3} \right )}}$

If D (a₁, b₁), E (a₂, b₂) and F (a₃, b₃) are the mid point of ΔABC, then its centroid is given by

$\\\mathrm{\mathbf{\left ( \frac{a_1+a_2+a_3}{3},\;\frac{b_1+b_2+b_3}{3} \right )}}$

Point of intersection of two lines -

Point of intersection of two lines

Equation of two non-parallel line is

$\\L_1=a_1x+b_1y+c_1=0\\L_2=a_2x+b_2y+c_2=0$

If P (x₁, y₁) is a point of intersection of L₁ and L₂ , then solving these two equations of the line by cross multiplication

$\frac{x_1}{b_1c_2-c_1b_2}=\frac{y_1}{c_1a_2-a_1c_2}=\frac{1}{a_1b_2-b_1a_2}$

We get,

$\mathbf{\left ( x_1,y_1 \right )=\left ( \frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1} \right )}$

Equation of Straight Line (Part 2) -

Equation of Straight Line

(c) Two-point form

The equation of a straight line passing through the two given points (x₁,y₁) and (x₁,y₁)is given by

The centroid of triangle ABC D(2,2)

Point of intersection P $\left ( -\frac{1}{5},\frac{2}{5} \right )$

equation of line DP is 8x – 11y + 6 = 0

Point (–9,–6) satisfies the equation

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avinash.dongre

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If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, $\frac{x}{1}= \frac{y}{4}= \frac{z}{5}$ is Q, then PQ is equal to :
Option: 1 $2\sqrt{42}$

Option: 5 $\sqrt{42}$

Option: 9 $6\sqrt{5}$

Option: 13 $3\sqrt{5}$

Intersection of line and plane -

Let the line

$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$ plane

$a_{1}x+b_{1}y+c_{1}z+d=0$ intersect at P

to find P assume general point on line as $\left ( x_{1}+\lambda a_{1}y_{1} +\lambda b_{1}z_{1}+\lambda c_{1}\right )$

now put it in plane to find $\lambda$ ,

$a_{1}\left ( x_{1}+\lambda a \right )+b_{1}\left ( y_{1}+\lambda b \right )+c_{1}\left ( z_{1}+\lambda c \right )+d=0$

Distance formula -

The distance between two points $A(x_{1},y_{1},z_{1})\, and \, B(x_{2},y_{2},z_{2})$ is = $\sqrt{\left ( x_{2}-x_{1} \right )^{2}+{\left ( y_{2}-y_{1} \right )^{2}}+{\left ( z_{2}-z_{1} \right )^{2}}}$

- wherein

$\vec{OA}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$

$\vec{OB}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$

$\underset{AB}{\rightarrow}$ = $\underset{OB}{\rightarrow}$ - $\underset{AB}{\rightarrow}$

$\underset{AB}{\rightarrow}$ $= \left ( x_{2}-x_{1} \right )\hat{i}+\left ( y_{2}-y_{1} \right )\hat{j}+\left ( z_{2}-z_{1} \right )\hat{k}$

$AB= \left | \underset{AB}{\rightarrow} \right |$

$AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$

PQ=2PM

Now, for finding M

$\frac{x}{1}=\frac{y}{4}=\frac{z}{5}=k(say)$

$\Rightarrow x=k,y=4k,z=5k$

Putting in equation of plane,

$k=\frac{11}{3}$

So, $M=\left (\frac{11}{3} ,\frac{44}{3},\frac{55}{3} \right )$

$\therefore PM=\sqrt{\frac{64}{9}+\frac{2500}{9}+\frac{2116}{9}}= 2\sqrt{130}$

$\Rightarrow PQ=4\sqrt{130}$

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vishal kumar

The line of intersection of the planes $\small \underset{r}{\rightarrow} .\left ( 3\hat{i} -\hat{j} +\hat{k}\right )= 1$ and $\small \underset{r}{\rightarrow} .\left ( \hat{i} +4 \hat{j} -2\hat{k}\right ) =2$ , is:
Option: 1 $-2 \hat{i}+7 \hat{j}+13 \hat{k}$
Option: 2 $2 \hat{i}+7 \hat{j}-13 \hat{k}$
Option: 3 $-2 \hat{i}-7 \hat{j}+13 \hat{k}$
Option: 4 $-2 \hat{i}+7 \hat{j}+13 \hat{k}$

As we have learned

Equation of line as intersection of two planes -

Let the two intersecting planes be

$ax+by+cz+d= 0$ and

$a_{1}x+b_{1}y+c_{1}z+d_{1}= 0$

then the parallel vector of line formed their intersection can be obtained by

$\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ a&b &c \\ a_{1} & b_{1} & c_{1} \end{vmatrix}= A\hat{i}+B\hat{j}+C\hat{k}(assumed)$

and points can be obtained by putting $z= 0$ and solving

$ax+by+d= 0$ and

$a_{1}x+b_{1}y+d_{1}= 0$ say $\alpha ,\beta$

Now the equation will be

$\frac{x-\alpha }{A}=\frac{y-\beta }{B}=\frac{z-0 }{C}$

$3x-y + z = 1 \\ and \: \: x + 4y - 2z = 2$

putting z = 0

3x-y = 1 and x + 4y = 2

$\Rightarrow 13 x = 6 \Rightarrow x = 6/13 \\ y = 5/3$

$vector \: \: along \: \: line = \begin{vmatrix} \hat i & \hat j &\hat k \\ 3 & -1 & 1\\ 1 & 4 &-2 \end{vmatrix} = -2 \hat i + 7 \hat j +13 \hat k$

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vishal kumar

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The tangent at the point (2, −2) to the curve, x²y²−2x=4(1−y) does not pass through the point :
Option: 1 (8,5)
Option: 2 (4,1/3)
Option: 3 (-2,-7)
Option: 4 (-4,-9)

As learnt in concept ABCD

$x^{2}y^{2}-2x=4(1-y)$

Differentiate both sides wrt.x.

$2xy^{y}+x^{2}.2yy{}'-2=4-4y{}'$

$y{}'=-\frac{(xy^{2}-1)}{2+x^{2}y}=\left ( \frac{1-xy^{2}}{2+x^{2}y} \right )$

at (2,-2) $y{}'=\frac{1-2\times (-2)^{2}}{2+2^{2}\times (-2)}=\frac{1-8}{2-8}=+\frac{7}{6}$

Equation of tangent is

$\frac{y+2}{x-2}=\frac{7}{6}=>7x-6y-26=0$

It doesn't pass through (-2,-7)

Concept ABCD

Slope of curve at a given point

To find slope, we differentiate with respect to x and find $\frac{dy}{dx}$

Eg in $x^{2}-y^{2}=4$

$\frac{dy}{dx}=\frac{x}{y}$

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vishal kumar

Let the normal at a point P on the curve $y^{2}-3x^{2}+y+10=0$ intersect the y-axis at $\left ( 0,\frac{3}{2} \right ).$ If m is the slope of the tangent at P to the curve, then $\left | m \right |$ is equal to
Option: 1 4
Option: 2 3
Option: 3 2
Option: 4 1

Equation of Straight Line (Part 1) -

Equation of Straight Line

(b) Point-Slope form

Let the equation of give line l with slope ‘m’ is

y = mx + c …..(i)

(x₁,y₁) lies on the line i

y₁= mx₁+c ……(ii)

From (i) and (ii) [(ii) - (i)]

y - y₁= m( x - x₁)

The equation of a straight line whose slope is given as ‘m’ and passes through the point (x₁,y₁) is .

$\begin{array}{c}{2 y y^{\prime}+y^{\prime}-6 x=0} \\ {y^{\prime}=\frac{6 x}{2 y+1}} \\ {\frac{-1}{y^{\prime}}=\frac{-(2 y+1)}{6 x}}(\text{slope of the normal})\end{array}$

Equation of the normal $y-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(x-x_{1}\right)$

Normal intersect at (0,3/2)

$\\\frac{3}{2}-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(0-x_{1}\right)\\ 8y_1-8=0\\ y_1=1\\ x_1=\pm2\\ |m|=4$

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Kuldeep Maurya

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Let the line $y=mx$ and the ellipse $2x^{2}+y^{2}=1$ intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at $\left ( -\frac{1}{3\sqrt{2}},0 \right )$ and $(0,\beta )$ , then $\beta$ is equal to :
Option: 1 $\frac{2}{\sqrt{3}}$
Option: 2 $\frac{2}{3}$
Option: 3 $\frac{2\sqrt{2}}{3}$
Option: 4 $\frac{\sqrt{2}}{3}$

Equation of Normal in Point Form and Parametric Form -

Equation of Normal in Point Form and Parametric Form

Point form:
$\\ {\text {The equation of normal at }\left(x_{1}, y_{1}\right) \text { to the ellipse, } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { is }} \\ {\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2}-b^{2}.}$

$\\ {\text { Let } P \text { be }\left(x_{1}, y_{1}\right)} \\\\ {\text { Equation of normal at } P \text { is } \frac{x}{2 x_{1}}-\frac{y}{y_{1}}=-\frac{1}{2}} \\\\ {\text { It passes through }\left(-\frac{1}{3 \sqrt{2}}, 0\right) \Rightarrow \frac{-1}{6 \sqrt{2} x_{1}}=-\frac{1}{2} \Rightarrow x_{1}=\frac{1}{3 \sqrt{2}}}$

$\\ {\text { So } y_{1}=\frac{2 \sqrt{2}}{3} \text { (as } P \text { lies in } 1 \text { 'quadrant) }} \\\\ {\text { So } \beta=\frac{y_{1}}{2}=\frac{\sqrt{2}}{3}}$

Correct option (4)

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Kuldeep Maurya

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is :

Option: 1 $2\sqrt{30}$
Option: 2 $\frac{7}{2}\sqrt{30}$
Option: 3 $3$
Option: 4 $3\sqrt{30}$

Shortest Distance between Two Lines -

Distance between two skew lines

If L₁ and L₂ are two skew lines, then there is one and only one line perpendicular to each of lines L1 and L2 which is known as the line of shortest distance.

Vector form

$\\\mathrm{\;\;\;L_1:\;\;\;\;}\vec{\mathbf r} &=\vec{\mathbf r}_{0}+\lambda \vec{\mathbf b} \\\\\mathrm{\;\;\;L_2:\;\;\;\;}\vec{\mathbf r} &=\vec{\mathbf r'}_{0}+\mu \vec{\mathbf b'}$

If $\overrightarrow{PQ}$ is the shortest distance vector between L₁ and L₂, then it being perpendicular to both $\vec{\mathbf b}$ and $\vec{\mathbf b'}$ , therefore, the unit vector $\hat{\mathbf n}$ along $\overrightarrow{PQ}$ would be

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\hat{\mathbf n}=\frac{\vec {\mathbf b}\times \vec{\mathbf b'}}{\left | \vec{\mathbf b}\times \vec{\mathbf b'} \right |}\\\mathrm{Then,\;\;\;\;\;\;\;\;\;\;\;\;}\overrightarrow{PQ}=\mathit{d}\hat{\mathbf n}$

where "d" is the magnitude of the shortest distance vector. Let θ be the angle between $\overrightarrow{ST}$ and $\overrightarrow{PQ}$ .

Then

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;PQ=ST\left |\cos\theta \right |}\\\\\text{but,}\;\;\;\;\;\;\;\;\;\;\;\;\;\cos \theta=\left|\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{ST}}}{|\overrightarrow{\mathrm{PQ}}||\overrightarrow{\mathrm{ST}}|}\right|\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\left|\frac{d \hat{n} \cdot\left(\vec{r'}_{0}-\vec{r}_{0}\right)}{d \;\mathrm{ST}}\right| \quad\left(\text { since } \overrightarrow{\mathrm{\;ST}}=\vec{r'}_{0}-\vec{r}_{0}\right)\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\left|\frac{\left(\vec{b} \times \vec{b'}\right) \cdot\left(\vec{r'}_{0}-\vec{r}_{0}\right)}{\mathrm{ST}\left|\vec{b} \times \vec{b'}\right|}\right|$

Hence, the required shortest distance is

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}d=\mathrm{PQ}=\mathrm{ST}|\cos \theta|\\\\\text{or}\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathbf d=\left|\frac{\left(\vec{\mathbf b} \times \vec{\mathbf b'}\right) \cdot\left(\vec{\mathbf r'}_{0} - \vec{\mathbf r}_{0}\right)}{\left|\vec{\mathbf b} \times \vec{\mathbf b'}\right|}\right|$

$\\\overrightarrow{\mathrm{AB}}=6 \hat{i}+15 \hat{j}+3 \hat{k}\\\vec p=3\hat i-\hat j+\hat k\\ \vec q=-3\hat i+2\hat j+4\hat k\\\vec p\times \vec q=\begin{vmatrix} \hat i &\hat j &\hat k \\ 3 &-1 &1 \\ -3 & 2 &4 \end{vmatrix}=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)\\\Rightarrow -6\hat i-15\hat j+3\hat k\\\text{shortest dist}=\frac{| \overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}$

= $3\sqrt{30}$

Correct Option (4)

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Kuldeep Maurya

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All Questions

If one end of a focal chord AB of the parabola is at then the equation of the tangent to it at B is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Shabareesh

The length of the minor axis (along y-axis) of an ellipse in the standard form is If this ellipse touches the line, then its eccentricity is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

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If are the eccentricities of the ellipse, and the hyperbola, respectively and is a point on the ellipse, then is equal to : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

Let C the centroid of the triangle with vertices Let P be the point of intersection of the lines and Then the line passing through the points C and P also passes through the point: Option: 1 Option: 2 Option: 3 Option: 4

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avinash.dongre

JEE Main high-scoring chapters and topics

If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, is Q, then PQ is equal to : Option: 1 Option: 5 Option: 9 Option: 13

If one end of a focal chord AB of the parabola $y^{2}=8x$ is at $A\left ( \frac{1}{2},-2 \right ),$ then the equation of the tangent to it at B is :
Option: 1 $x+2y+8=0$
Option: 2 $2x-y-24=0$
Option: 3 $x-2y+8=0$
Option: 4 $2x+y-24=0$

If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, $\frac{x}{1}= \frac{y}{4}= \frac{z}{5}$ is Q, then PQ is equal to :
Option: 1 $2\sqrt{42}$

Option: 5 $\sqrt{42}$

Option: 9 $6\sqrt{5}$

Option: 13 $3\sqrt{5}$

The tangent at the point (2, −2) to the curve, x²y²−2x=4(1−y) does not pass through the point :
Option: 1 (8,5)
Option: 2 (4,1/3)
Option: 3 (-2,-7)
Option: 4 (-4,-9)

Let the normal at a point P on the curve $y^{2}-3x^{2}+y+10=0$ intersect the y-axis at $\left ( 0,\frac{3}{2} \right ).$ If m is the slope of the tangent at P to the curve, then $\left | m \right |$ is equal to
Option: 1 4
Option: 2 3
Option: 3 2
Option: 4 1

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is :

Option: 1 $2\sqrt{30}$
Option: 2 $\frac{7}{2}\sqrt{30}$
Option: 3 $3$
Option: 4 $3\sqrt{30}$