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    RD Sharma Class 12 Exercise MCQ The Plane Solutions Maths - Download PDF Free Online

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    RD Sharma Class 12 Exercise MCQ The Plane Solutions Maths - Download PDF Free Online

    Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:52 AM IST

    When it comes to preparing for board exams, RD Sharma books are the best. For many students, board exams have always been a ruse. Still, when they refer to this book in combination with Rd Sharma Class 12th Exercise MCQ solutions, they find it much easier to relax and understand the concept in a much better manner. RD Sharma solutions This exercise contains 24 questions which are based on the Equation of the plane containing two lines, Equation of the plane in scalar product form, Reflection of point, Vector and Cartesian form, equation in normal form and distance between the plane.

    RD Sharma Class 12 Solutions Chapter28 MCQ The Plane - Other Exercise

    The Plane Excercise: MCQ

    The Plane exercise multiple choice question 1

    Answer:
    Option (a).
    Hint:
    Simplify the given equation.
    Given:
    2x-(1+\lambda )y+3\lambda z=0
    Solution:
    The given equation plane is,
    2x-(1+\lambda )y+3\lambda z=0
    We can rewrite the equation of the given plane as,
    2x-(1+\lambda )y+3\lambda z=0
    2x-y-\lambda (y-3z)=0

    So, the given plane passes through the intersection of the planes

    2x-y=0 \text { and }y-3z=0.

    The Plane exercise multiple choice question 2

    Answer:
    Option(b)
    Hint:
    Angle between two planes is 0.
    Given:
    2x-y+z=6 \text { and } x+y+2z=3
    Solution:
    We know, the angle between two planes
    a_1x+b_1y+c_1z+d_1=0 \text { and } a_2x+b_2y+c_2z+d_2=0 \text { is }
    \theta =cos^{-1}\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1\, ^2+b_1\, ^2+c_1\, ^2}.\sqrt{a_2\, ^2+b_2\, ^2+c_2\, ^2}}
    Here a1 = 2; b1 = -1; c1 = 1; d1 = -6 and a2 = 1; b2 = 1; c2 = 2; d2 = -3
    So, the acute angle between the planes
    2x-y+z=6 \text { and } x+y+2z=3is
    \begin{aligned} &\theta =cos^{-1}\frac{((2.1)+(-1.1)+(1.2))}{\sqrt{2^2+(-1)^2+1^2}.\sqrt{1^2+1^2+2^2}}\\ &=cos^{-1}\frac{(2-1+2)}{\sqrt{4+1+1}.\sqrt{1+1+4}}\\ &=cos^{-1}\frac{3}{\sqrt{6}.\sqrt{6}}\\ &=cos^{-1}\frac{3}{6}\\ &\theta =60^{o} \end{aligned}
    Therefore, the acute angle between the planes
    2x-y+z=6 \text { and } x+y+2z=3 \text { is } 60^o

    The Plane exercise multiple choice question 3

    Answer:
    Option (d)
    Hint:
    \lambda is a scalar quantity.
    Given:
    x+2y+3z=4 \text { and }2x+y-z=-5
    are perpendicular to the plane
    5x+3y+6z+8=0
    Solution:
    The equation of the plane through the intersection of the planes
    x+2y+3z=4 \text { or } x+2y+3x-4=0 \text { and }2x+y-z=-5 \text { or } 2x+y-z+5=0
    is given as,
    \begin{aligned} &(x+2y+3z-4)-\lambda (2x+y-z+5)=0\\ &x(1+2\lambda )+y(2+\lambda )+z(3-\lambda )-4+5\lambda =0, \end{aligned}
    where \begin{aligned} &\lambda \end{aligned} is a scalar.
    Given, that the required plane is perpendicular to the plane
    \begin{aligned} &5x+3y+6z+8=0 \end{aligned}
    so we have,
    \begin{aligned} &5(1+2\lambda )+3(2+\lambda )+6(3-\lambda )=0\\ &5+10\lambda +6+3\lambda +18-6\lambda =0\\ &29+7\lambda =0\\ &\lambda =-\frac{29}{7} \end{aligned}
    Therefore, the equation of the required plane is,
    \begin{aligned} &(x+2y+3z-4)-\frac{29}{7}(2x+y-z+5)=0\\ &7(x+2y+3z-4)-29(2x+y-z+5)=0\\ &7x+14y+21z-28-58x-29y+29z-145=0\\ &-51x-15y+50z-173=0\\ &51x+15y-50z+173=0 \end{aligned}

    The Plane exercise multiple choice question 4

    Answer:
    Option (c).
    Hint:
    Distance between two parallel planes is 0.
    Given:
    2x+2y-z=0 \text { and } 4x+4y-2z+5=0
    Solution:
    We know that, the distance between two parallel planes:
    Ax+By+Cz+D_1=0 \qquad \qquad \dots(1)
    and
    Ax+By+Cz+D_2=0 \qquad \qquad \dots(2)
    is given by
    D=\frac{\left | D_2-D_{1} \right |}{\sqrt{A^2+B^2+C^2}}
    Here, the two parallel planes are given as:
    2x+2y-z+2=0 \qquad \qquad \dots (3)
    and
    4x+4y-2z+5=0
    i.e.
    2x+2y-z+\frac{5}{2}=0 \qquad \qquad \dots (4)
    Compiling equation (3) with equation (1), and equation (4) with equation (2), we get;
    A = 2,B=2,C=-1,D_1=2,D_2=\frac{5}{2}.
    So, the distance between the given two parallel planes are,
    \begin{aligned} &D=\frac{\left | \frac{5}{2}-2 \right |}{\sqrt{2^2+2^2+(-1)^2}}\\ &D=\frac{\frac{1}{2}}{\sqrt{4+4+1}}\\ &D=\frac{1}{2\sqrt{9}}\\ &D=\frac{1}{2\times 3}\\ &D=\frac{1}{6} \end{aligned}
    Hence, the distance between the parallel planes
    2x+2y-z=0 \text { and } 4x+4y-2z+5=0 \text { is } \frac{1}{6}.

    The Plane exercise multiple choice question 5

    Answer:
    Option (b)
    Hint:
    Solve the equation simultaneously.
    Given:
    (1, 3, 4) in the plane 2x - y + z + 3 = 0
    Solution:
    We know, if the image of a point P(x0, y0, z0) on a plane Ax + By + Cz + D = 0 . . . (1) is Q(x1, y1, z1) then

    The given plane is 2x - y + z + 3 = 0 . . . . .(2)
    By comparing equations (2) and (3)
    A = 2, B = -1, C = 1, D = 3
    And here x_0 = 1, y_0 = 3, z_0 = 4
    So
    \begin{aligned} &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2[(2\times 1)+((-1)\times 3)+(1\times 4)]}{2^2+(-1)^2+1^2}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2(2-3+4+3)}{4+1+1}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2\times 6}{6}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=-2 \end{aligned}
    Therefor,
    x1 = -4 + 1
    =-3
    y1 = 2 + 3
    =5
    z1 = -2 + 4
    =2
    So, the image of the (1, 3, 4), in the plane 2x - y + 3 = 0 , is (-3, 5, 2).

    The Plane exercise multiple choice question 6

    Answer:
    Option (d).
    Hint:
    See for co-planer.
    Given:
    \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-0}{3} \text { and } \frac{x-0}{-2}=\frac{y-2}{-3}=\frac{z+1}{-1}
    Solution:
    We know, the two lines given as,
    \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-0}{3} \text { and } \frac{x-0}{-2}=\frac{y-2}{-3}=\frac{z+1}{-1}
    \begin{aligned} &\vec{b_1}=2\hat{i}-\hat{j}+3\hat{k}\\ &\vec{b_2}=-\hat{i}-3\hat{j}-\hat{k}\\ &=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 2 &-1 &3 \\ -1 &-3 &-1 \end{vmatrix}\\ &=-8\hat{i}-\hat{j}+5\hat{k} \end{aligned}
    Lines passes through (1,-1,0)
    \begin{aligned} &(x-x_1)a+(y-y_1)b+(z-z_1)c=0\\ &(x-1)(-8)+(y+1)(-1)+(z-0)5=0\\ &-8x+8-y-1+5z=0\\ &-8x-y+5z+7=0\\ &8x+y-5z-7=0 \end{aligned}
    Hence, option a is correct.

    The Plane exercise multiple choice question 7

    Answer:
    Option (a)
    Hint:
    Use cross product of vector.
    Given:
    \vec{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})
    Solution:
    The given plane is
    \vec{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})
    So, it is clear from the given equation of plane, that the plane passing through a point (\hat{i}-\hat{j}) and parallel to the two vectors (\hat{i}+\hat{j}+\hat{k}) \text { and } (\hat{i}-2\hat{j}+3\hat{k}).
    Therefore, the equation of the vector normal to the plane is given as
    \begin{aligned} &\widehat{n}=[(\hat{i}+\hat{j}+\hat{k}) \times (\hat{i}-2\hat{j}+3\hat{k})]\\ &=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &1 &1 \\ 1 &-2 &3 \end{vmatrix}\\ &=\left [ \left \{ (1\times 3)-(-2\times 1) \right \}\widehat{i}-\left \{ (1\times 3)-(1\times 1) \right \}\widehat{j}+\left \{ (1\times -2)-(1\times 1)\widehat{k} \right \} \right ]\\ &=5\widehat{i}-2\widehat{j}-3\widehat{k} \end{aligned}
    So, in scalar product form the vector equation of the plane is given as,
    \begin{aligned} &(\overrightarrow{r}-( \hat{i}-\hat{j})).\widehat{n}=0\\ &(\overrightarrow{r}-( \hat{i}-\hat{j})).(5\widehat{i}-2\widehat{j}-3\widehat{k})=0\\ &\overrightarrow{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=5+2\\ &\overrightarrow{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=7 \end{aligned}

    The Plane exercise multiple choice question 8

    Answer:
    Option (b)
    Hint:
    Use scalar product of vector
    Given:
    \overrightarrow{r}=2\widehat{i}-2\widehat{j}+3\widehat{k}+\lambda (\widehat{i}+\widehat{j}+4\widehat{k})
    from the plane
    \overrightarrow{r}.(\widehat{i}+5\widehat{j}+\widehat{k})=5
    Solution:
    We have the straight line given as,
    \overrightarrow{r}=2\widehat{i}-2\widehat{j}+3\widehat{k}+\lambda (\widehat{i}+\widehat{j}+4\widehat{k})
    And the plane as,
    \begin{aligned} &\overrightarrow{r}.(\widehat{i}+5\widehat{j}+\widehat{k})=5\\ &x-5y+2=5\\ &\text { i.e. }x-5y+2-5=0 \end{aligned}
    the normal vector of the plane given as,
    \overrightarrow{n}=(\widehat{i}-5\widehat{j}+\widehat{k})
    if the straight line and the plane are parallel,
    scalar product will be zero.
    \begin{aligned} &(\widehat{i}+\widehat{j}+4\widehat{k}).(\widehat{i}-5\widehat{j}+\widehat{k})=1+[1\times (-5)]+(4\times 1)\\ &=1-5+4\\ &=0 \end{aligned}
    Hence, the point (2, -2, 3) is on the straight line. Distance from point (2, -2, 3) to the plane will be equal to the distance of the line from the plane. We know, that the distance of a point (x0, y0, z0) from plane Ax + By + Cz + D = 0 ...(2) is
    \begin{aligned} &=\frac{\left | Ax_0+By_0+Cz_0+D \right |}{\sqrt{A^2+B^2+C^2}} \end{aligned}
    On comparing, equation (1), i.e., x - 5y + z - 5 = 0 with equation (2), we get,
    A = 1, B = -5, C = 1, D = -5
    So, the distance is,
    \begin{aligned} &=\frac{\left [ (1\times 2)+(-5\times -2)+(1\times 3)+(-5) \right ]}{\sqrt{1^2+(-5)^2+1^2}}\\ &=\frac{\left | 2+10+3-5 \right |}{\sqrt{1+25+1}}\\ &=\frac{10}{\sqrt{27}}\\ &=\frac{10}{3\sqrt{3}} \end{aligned}

    The Plane exercise multiple choice question 9

    Answer:
    Option (a)
    Hint:
    Solve the given equations simultaneously.
    Given:
    x+y+z+3=0,\: 2x-y+3z+1=0 \text { and } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}
    Solution:
    Equation of line passing through the line
    x+y+z+3=0,\: 2x-y+3z+1=0
    is given by,
    \begin{aligned} &x+y+z+3+k(2x-y+3z+1)=0 \qquad \qquad \dots (1)\\ &x(1+2k)+y(1-k)+z(1+3k)+3+k=0 \end{aligned}
    where k is constant
    Again, the required plane is parallel to the line
    \begin{aligned} &\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \end{aligned}
    So, we have
    \begin{aligned} &\left [ 1\times (1+2k) \right ]+\left [ 2\times (1-k) \right ]+\left [ 3\times (1+3k) \right ]=0\\ &1+2k+2-2k+3+9k=0\\ &9k=-6\\ &k=-\frac{2}{3} \end{aligned}
    Put k in eq. (1)
    \begin{aligned} &(x+y+z+3)-\frac{-2}{3}(2x-y+3z+1)=0\\ &3(x+y+z+3)-2(2x-y+3z+1)=0\\ &3x+3y+3z+9-4x+2y-6z-2=0\\ &x-5y+3z=7 \end{aligned}
    Therefore, the equation of the plane through the line x + y + z + 3 = 0, 2x - y + 3z + 1 = 0 and the parallel to the line
    \frac{x}{1}=\frac{y}{2}=\frac{z}{3}
    x - 5y + 3z = 7

    The Plane exercise multiple choice question 10

    Answer:
    Option (a)
    Hint:
    Put \lambda = 1
    Given:
    \overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})
    Solution:
    The plane contains the line
    \overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})
    so, the plane contains the point
    (\widehat{i}+2\widehat{j}+3\widehat{k})
    Put \lambda = 1
    -2\widehat{i}+3\widehat{j}+4\widehat{k}+3\widehat{i}-2\widehat{j}-\widehat{k} \text { i.e } (\widehat{i}-5\widehat{j}+3\widehat{k})
    so, we got the plane, they are:
    (\widehat{i}+2\widehat{j}+3\widehat{k}); (-2\widehat{i}-3\widehat{j}+4\widehat{k}); \text { and } (\widehat{i}-5\widehat{j}+3\widehat{k})
    Let,
    \begin{aligned} &\overrightarrow{a}=(\widehat{i}-5\widehat{j}+3\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k}) \text { and }\\ &\overrightarrow{b}=(\widehat{i}-5\widehat{j}+3\widehat{k})-(-2\widehat{i}-3\widehat{j}+4\widehat{k})\\ &\text { So, }\overrightarrow{a}=-7\widehat{j}\text { and }\overrightarrow{b}=3\widehat{i}-2\widehat{j}-\widehat{k} \end{aligned}

    The normal of two vectors \begin{aligned} &\overrightarrow{a} \end{aligned} and \begin{aligned} &\overrightarrow{b} \end{aligned}

    \begin{aligned} &\overrightarrow{a}=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 0 &-7 &0 \\ 3 &-2 &1 \end{vmatrix}\\ &=\left [ \left \{ (-7)\times (-1)-((-2)\times 0) \right \}\widehat{i}+\left \{ (0\times (-1))-(0\times (-2)) \right \}\widehat{j}+\left \{ (0\times (-2))-((-7)\times 3) \right \}\widehat{k} \right ]\\ &=7\widehat{i}+21\widehat{k} \end{aligned}
    The general equation of plane is,
    \begin{aligned} &\left [ (x-1)\widehat{i}+(y-2)\widehat{j}+(z-3)\widehat{k} \right ].(7\widehat{i}+21\widehat{k})=0\\ &7(x-1)+21(z-3)=0\\ &7x-7+21z-63=0\\ &7x+21z=70\\ &x+3z=10\\ &\text { or }\\ &\overrightarrow{r}.(\widehat{i}+3\widehat{k})=10, \qquad \left [ \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k} \right ] \end{aligned}
    Hence, the vector equation of the plane containing the line
    \overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})
    and of the point
    (\widehat{i}+2\widehat{j}+3\widehat{k})
    is
    \begin{aligned}&\overrightarrow{r}.(\widehat{i}+3\widehat{k})=10 \end{aligned}

    The Plane exercise multiple choice question 11

    Answer:
    Option (c)
    Hint:
    Use the centroid formula.
    Given:
    \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=k
    Solution:
    A plane meets the co-ordinate area of (A, B, C) such that the centroid of \Delta ABC is the point (a,b,c).
    Let the co-ordinate of the point
    A(\alpha , 0, 0),\: B(0,\beta ,0),\: C(0,0,\gamma )
    By the centroid formula,
    \begin{aligned} &a =\frac{\alpha +0+0}{3}\\ &\Rightarrow \alpha =3a\\ &b=\frac{0+\beta +0}{3}\\ &\Rightarrow \beta =3b\\ &c=\frac{0+0+\gamma }{3}\\ &\Rightarrow \gamma =3c \end{aligned}
    The Intercept form of the plane–
    \frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1
    If the plane makes intercepts at (p, 0, 0), (0, q, 0) and (0, 0, r) with the axes x, y, z respectively.
    Here,
    p=\alpha =3a,\: q=\beta =3b,\: r=\gamma =3c
    So, the equation of the plane is,
    \begin{aligned} &\frac{x}{3a}+\frac{y}{3b}+\frac{z}{3c}=1\\ &\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3 \qquad \qquad \qquad \dots(1) \end{aligned}
    By comparing the equation (1) with given equation of the plane
    \begin{aligned} &\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=k \end{aligned},
    we get k = 3.

    The Plane exercise multiple choice question 12

    Answer:
    Option (c)
    Hint:
    The plane, must satisfy both the equation.
    Given:
    \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}
    Solution:
    Let, the point of intersection of the line
    \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}
    And the plane x + y + z = 17 be (x0, y0, z0) So, (x0, y0, z0) satisfy the both equations,
    \begin{aligned} &\frac{x_0-3}{1}=\frac{y_0-4}{2}=\frac{z_0-5}{2}=k, \text { (Let) }\\ &\text { i.e. }x_0=k+3\\ &y_0=2k+4\\ &z_0=2k+5 \end{aligned}
    Put these values in the equation of plane x + y + z = 17
    \begin{aligned} &(k+3)+(2k+4)+(2k+5)=17\\ &5k+12=17\\ &k=1\\ &\therefore x_0=1+3\\ &=4\\ &y_0=2\times 1+4\\ &=6\\ &z_0=2\times 1+5\\ &=7\\ \end{aligned}
    Hence, the distance between point (4, 6, 7) and (3, 4, 5) is,
    \begin{aligned} &=\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}\\ &=\sqrt{1^2+2^2+2^2}\\ &=\sqrt{1+4+4}\\ &=3 \end{aligned}

    The Plane exercise multiple choice question 13

    Answer:
    -2\widehat{i}+7\widehat{j}+13\widehat{k}
    Hint:
    Solve linear equations in two variables
    Given:
    \overrightarrow{r}.(3\widehat{i}-\widehat{j}+\widehat{k})=1 \text { and } \overrightarrow{r}.(\widehat{i}-4\widehat{j}-2\widehat{k})=2
    Solution:
    The two planes are
    \overrightarrow{r}.(3\widehat{i}-\widehat{j}+\widehat{k})=1 \text { and } \overrightarrow{r}.(\widehat{i}-4\widehat{j}-2\widehat{k})=2
    The line of intersection of planes
    \overrightarrow{r}.(3\widehat{i}-\widehat{j}+\widehat{k})=1 \text { and } \overrightarrow{r}.(\widehat{i}-4\widehat{j}-2\widehat{k})=2
    Can be written as 3x - y + z = 1 ……………… (i)
    x + 4y - 2z = 2 …………… (ii)
    Solving eqn (i) & (ii)
    \frac{x}{-2}=\frac{y}{7}=\frac{z}{13}
    Vector equation is
    -2\widehat{i}+7\widehat{j}+13\widehat{k}

    The Plane exercise multiple choice question 14

    Answer:
    Option (c)
    Hint:
    Calculate the perpendicular distance of the plane from the origin.
    Given:
    \frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}
    Solution:
    Let, the equation of the plane be Ax + By + Cz + D = 0, as the plane is perpendicular to, so
    We have,
    A = 3, B = 0 and C = 4 as the plane passes through (1, 1, 1), we have,
    \begin{aligned} &(A\times 1)+(B\times 1)+(C\times 1)+D=0\\ &A+B+C+D=0\\ &3+0+4+D=0\\ &D=-7 \end{aligned}
    So, the equation of the plane becomes
    3x + 4z - 7 = 0
    Now, the perpendicular distance of the plane from the origin is
    \begin{aligned} &\frac{\left | Ax_0+By_0+Cz_0+D \right |}{\sqrt{A^2+B^2+C^2}}=\frac{\left | (3\times 0)+(0\times 0)+(4\times 0)-7 \right |}{\sqrt{3^2+0^2+4^2}}\\ &\left [ \because (x_0, y_0,z_0)\approx (0,0,0) \right ]\\ &=\frac{\left | 0-7 \right |}{\sqrt{9+16}}\\ &=\frac{\left | -7 \right |}{\sqrt{25}}\\ &=\frac{7}{5} \end{aligned}

    The Plane exercise multiple choice question 15

    Answer:
    Option (a)
    Hint:
    Use straight line & vector cross product.
    Given:
    x - 1 = 2y - 5 = 2z and 3x = 4y -11 = 3z - 4
    Solution:
    The required plane is parallel to the lines
    x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z - 4
    Equation of the lines can be written as,
    \frac{x-1}{1}=\frac{2y-5}{1}=\frac{2z}{1}
    or
    \frac{x-1}{1}=\frac{y-\frac{5}{2}}{\frac{1}{2}}=\frac{z}{\frac{1}{2}}
    and,
    \frac{3x}{1}=\frac{4y-11}{1}=\frac{3z-4}{1}
    or,
    \frac{x}{\frac{1}{3}}=\frac{y-\frac{11}{4}}{\frac{1}{4}}=\frac{z-\frac{4}{3}}{\frac{1}{3}}=\mu , \text { Let }
    So, we know the straight lines as
    \begin{aligned} &(1+\lambda )\widehat{i}+\left ( \frac{5}{2}+\frac{1}{2}\lambda \right )\widehat{j}+\frac{1}{2}\widehat{k}=0 \text { or, }\\ &\widehat{i}+\frac{5}{2}\widehat{j}+\lambda \left ( \widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k} \right )=0 \end{aligned}
    And,
    \begin{aligned} &\frac{1}{3}\mu \widehat{i}+\left ( \frac{11}{4}+\frac{1}{4}\mu \right )\widehat{j}+\left ( \frac{4}{3}+\frac{1}{3}\mu \right )\widehat{k}=0, \text { or }\\ &\frac{11}{4}\widehat{j}+\frac{4}{3}\widehat{k}+\mu \left ( \frac{1}{3}\widehat{i}+\frac{1}{4}\widehat{j}+\frac{1}{3}\widehat{k} \right )=0 \end{aligned}
    We have the normal vector of the plane as,
    \begin{aligned} &\overrightarrow{n}=\left (\widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k} \right )\times \left ( \frac{1}{3}\widehat{i}+\frac{1}{4}\widehat{j}+\frac{1}{3}\widehat{k} \right )\\ &=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ \\ 1 &\frac{1}{2} &\frac{1}{2} \\ \\ \frac{1}{3} &\frac{1}{4} &\frac{1}{3} \end{vmatrix}\\ &=\left ( \left ( \frac{1}{2}\times \frac{1}{3} \right )-\left ( \frac{1}{2}\times \frac{1}{4} \right ) \right )\widehat{i}-\left ( \left ( 1\times \frac{1}{3} \right )-\left ( \frac{1}{2}\times \frac{1}{3} \right ) \right )\widehat{j}+\left ( \left ( 1\times \frac{1}{4} \right )-\left (\frac{1}{2}\times \frac{1}{3} \right ) \right )\widehat{k}\\ &=\left ( \frac{1}{6}-\frac{1}{8} \right )\widehat{i}-\left ( \frac{1}{3}-\frac{1}{6} \right )\widehat{j}+\left ( \frac{1}{4}-\frac{1}{6} \right )\widehat{k}\\ &=\left (\frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right ) \end{aligned}
    So, the equation of plane is
    \begin{aligned} &(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0\\ &\text { Where } \overrightarrow{a}=2\widehat{i}+3\widehat{j}+3\widehat{k},\\ &\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}\\ &\left ( x\widehat{i}+y\widehat{j}+z\widehat{k} \right ).\overrightarrow{n}=(2\widehat{i}+3\widehat{j}+3\widehat{k}).\overrightarrow{n}\\ &\left ( x\widehat{i}+y\widehat{j}+z\widehat{k} \right ).\left ( \frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right )=(2\widehat{i}+3\widehat{j}+3\widehat{k}).\left ( \frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right ) \end{aligned}
    x - 4y + 2z + 4 = 0, (by solving further)

    The Plane exercise multiple choice question 16

    Answer:
    Option (b)
    Hint:
    Use simultaneous equation.
    Given:
    (-1, -5, -10)
    Solution:
    Let the point of intersection of the line
    \overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k}) -1, -5
    And the plane
    \overrightarrow{r}.(\widehat{i}-\widehat{j}+\widehat{k})=5 \text { be } \left (x_0,y_0,z_0 \right )
    As x0, y0, z0 is the point of intersection of the line and the plane, so the position vector of this point i.e.,
    \overrightarrow{r_0}.(x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k})
    Must satisfy both equations of line and the equation of plane.
    Substituting \overrightarrow{r_0} in plane of \overrightarrow{r} in both the equations, we get,
    (x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k})=(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k})
    and
    (x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k}).(\widehat{i}-\widehat{j}+\widehat{k})=5 \qquad \qquad \dots (2)
    Substituting, these values in equation (2)
    \begin{aligned} &\left [ ((2+3\lambda )\times 1)-(1\times (-1+4\lambda )) \right ]+\left [ 1\times (2+12\lambda ) \right ]=5\\ &2+3\lambda +1-4\lambda +2+12\lambda =5\\ &11\lambda =0\\ &\lambda =0 \end{aligned}
    Therefore,
    \begin{aligned} &x_0=2+3\lambda \\ &=2,\\ &y_0=-1+4\lambda \\ &=-1\\ &z_0=2+12\lambda \\ &=2 \end{aligned}
    Hence, the point of intersection is (2, -1, 2)
    Now, the distance between the point (-1, -5, -10) and (2, -1, 2) is
    \begin{aligned} &=\sqrt{[2-(-1)]^2+[(-1)-(-5)]^2+[2-(-10)]^2}\\ &=\sqrt{3^2+4^2+12^2}\\ &=\sqrt{169}\\ &=13 \end{aligned}
    Hence, the required distance between the point (-1, -5, -10) and the point where the line
    \begin{aligned} &\overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k}) \end{aligned}
    And the plane
    \begin{aligned} &\overrightarrow{r}.(\widehat{i}-\widehat{j}+\widehat{k})=5 \end{aligned}
    intersects is 13 units

    The Plane exercise multiple choice question 17

    Answer:
    Option (a)
    Hint:
    \lambda is a scalar.
    Given:
    ax + by + cz + d = 0
    Solution:
    The equation of the plane through the intersection of the planes
    ax + by + cz + d = 0 and lx + my + nz + p = 0
    Is given by
    (ax + by + cz + d) + (lx + my + nz + p) = 0, [ where \lambda is a scalar ]
    x (a + l \lambda) + y (b + m \lambda) + z (c + n \lambda) + d + p \lambda = 0
    Given that the required plane is parallel to the lines y = 0, z = 0, i.e., x-axes so, we have
    1 (a + l \lambda) + 0 ( b + m \lambda) + 0 (c + n \lambda) = 0
    a + l \lambda = 0
    \lambda =-\frac{a}{l}
    (a + l \lambda )x + (b + m \lambda) y + (c + n\lambda) z + (d + pλ ) = 0
    \begin{aligned} &\left ( a+l\times \frac{-a}{l} \right )x +\left ( b+m\times \frac{-a}{l} \right )y+\left ( c+n\times \frac{-a}{l} \right )z+\left ( d+p\times \frac{-a}{l} \right )=0 \end{aligned}
    (b l - a m) y + (c l - a n) z + d l - a p = 0
    Hence, option a is correct.

    The Plane exercise multiple choice question 18

    Answer:
    Option (a)
    Hint:
    Use simultaneous equation.
    Given:
    The equation of the plane which cuts equal intercepts of unit length of the co-ordinate axes.
    Solution:
    We know that the general equation of a planes
    Ax+By+Cz+D=0, \text { where } D\neq 0 \qquad \qquad \dots(1)
    Here, A, B, C are the coordinate of a normal vector to the plane, while (x, y, z) are the coordinates of any point through which the plane passes.
    Again, we know the intercept form of plane
    \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \qquad \qquad \dots(2), \text { where }A=\frac{D}{a},B=-\frac{D}{b},C=-\frac{D}{c} A
    And the plane makes intercepts at (a, 0, 0), (0, b, 0) and (0, 0, c) with the axes x, y, z respectively.
    Here a = b = c = 1
    Put in the equation (2)
    \frac{x}{1}+\frac{y}{1}+\frac{z}{1}=1
    Hence the equation is x + y + z = 1

    The Plane exercise multiple choice question 19

    Answer:
    Option (d)
    Hint:
    Solve the equation to get the values of x, y and z
    Given:
    xy + yz = 0
    Solution:
    We have
    xy + yz = 0
    x (y + z) = 0
    therefore x = 0, y + z = 0
    Above one equation of planes normal to the plane x = 0 is \widehat{i}
    y + z = 0 is
    \widehat{j}+\widehat{k},
    And normal to the plane
    \widehat{i}.(\widehat{j}+\widehat{k})=0
    So, planes are perpendicular.

    The Plane exercise multiple choice question 20

    Answer:
    Option (d)
    Hint:
    See the reflection with the graph
    Given:
    (\alpha ,\beta ,\gamma )
    Solution:
    With the help of graph,
    In xy-plane, the reflection of the point (\alpha ,\beta ,\gamma ) is (\alpha ,\beta ,-\gamma )

    The Plane exercise multiple choice question 21

    Answer:
    Option (c)
    Hint:
    Use vector dot product for direction.
    Given:
    2x - 3y + 6z - 11 = 0
    Solution:
    We have equation of plane are 2x - 3y + 6z - 11 = 0
    Normal to the plane is
    \overrightarrow{n} = 2\widehat{i}-3\widehat{j}+6\widehat{k}
    Also x-axis is along the vector
    \overrightarrow{a} = \widehat{i}+0\widehat{j}+0\widehat{k}
    According to the question
    \begin{aligned} &sin \alpha =\frac{\left | \overrightarrow{a}.\overrightarrow{n} \right |}{\left | \overrightarrow{a} \right |.\left | \overrightarrow{n} \right |}\\ &=\frac{\left | \widehat{i}.(2\widehat{i}-3\widehat{j}+6\widehat{k}) \right |}{\sqrt{1}.\sqrt{4+9+36}}\\ &=\frac{2}{7} \end{aligned}

    The Plane exercise multiple choice question 22

    Answer:
    Option (d)
    Hint:
    Find the angle with vector dot product.
    Given:
    \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}
    Solution:
    Given line
    \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}
    is parallel to the vector
    \overrightarrow{b}=3\widehat{i}+4\widehat{j}+5\widehat{k}
    Equation to the plane is
    2x - 2y + z = 5
    Normal to the plane is
    \overrightarrow{n}=2\widehat{i}-2\widehat{j}+\widehat{k}
    If angle between line and plane is \theta
    Then
    \begin{aligned} &sin\theta =\frac{\left | \overrightarrow{b}.\overrightarrow{n} \right |}{\left | \overrightarrow{b} \right |.\left | \overrightarrow{n} \right |}\\ &=\frac{\left | (3\widehat{i}+4\widehat{j}+5\widehat{k}).(2\widehat{i}-2\widehat{j}+\widehat{k}) \right |}{\sqrt{3^2+4^2+5^2.\sqrt{4+4+1}}}\\ &=\frac{\left | 6-8+5 \right |}{\sqrt{50}.\sqrt{9}}\\ &=\frac{3}{15\sqrt{3}}\\ &=\frac{1}{5\sqrt{3}}\\ &\therefore sin\theta =\frac{\sqrt{2}}{10} \end{aligned}

    The Plane exercise multiple choice question 23

    Answer:
    Option (a)
    Hint:
    Find magnitude of n vector.
    Given:
    \overrightarrow{r}.\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )=1
    Solution:
    \overrightarrow{r}.\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )=1
    Let,
    \begin{aligned} &\overrightarrow{n}=\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )\\ &\left | \overrightarrow{n} \right |=\sqrt{\left ( \frac{2}{7} \right )^2+\left ( \frac{3}{7} \right )^2+\left ( -\frac{6}{7} \right )^2}\\ &=1 \end{aligned}
    N is a unit vector.

    The Plane exercise multiple choice question 24

    Answer:
    Option (b)
    Hint:
    Use vector addition.
    Given:
    (-1, 5, 4)
    Solution:

    Line passes through (-1, 5, 4) and perpendicular to the plane z = 0

    Equation of plane

    0. x + 0. y + 1. z = 0
    Normal of plane (0, 0, 1)
    Equation of line
    r = (-1, 5, 4) + \lambda (0, 0, 1)
    \therefore \overrightarrow{r}=-\widehat{i}+5\widehat{j}+4\widehat{k}+\lambda \widehat{k}


    Rd Sharma Class 12 Chapter 28 Exercise MCQ is organised in a stepwise manner with hints for complex problems so that a student can find the best path to solve them. The advantages of these solutions are as follows:

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    RD Sharma Chapter-wise Solutions

    Frequently Asked Question (FAQs)

    1. What is the definition of a Plane Figure?

    Any geometric figure with no thickness is referred to as a plane figure.

    2. Is the RD Sharma textbook a good choice for class 12?

    The CBSE recommends RD Sharma as a reference book. RD Sharma's class 12 solutions include numerous solved illustrations, pictorial representations, and more than one method for problem-solving. It also teaches students mathematical tricks and concepts that help them solve problems quickly and accurately. It is an excellent choice for practice questions.

    3. Is R S Agrawaal superior to RD Sharma?

    RD Sharma explains the concepts more clearly. It also includes a large number of practise questions of various types. As a result, it prepares students to perform better in exams.

    4. What is the main concept of the chapter ‘The Plane’?

    This chapter explains the plane concept of geometry.

    5. What is the best website for studying RD Sharma Solutions for Class 12 Maths?

    On the Career360 website, you can find Rd Sharma Class 12th Exercise MCQ solutions, as well as step-by-step answers to all of the questions in the RD Sharma textbook. As a result, in order to understand the important topics, students in Class 12 should learn all of the concepts covered in the syllabus.

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    Individuals following a career as health inspectors have to face resistance and lack of cooperation while working on the sites. The health inspector's job description includes taking precautionary measures while inspecting to save themself from any external injury and the need to cover their mouth to avoid toxic substances. A health inspector does the desk job as well as the fieldwork. Health inspector jobs require one to travel long hours to inspect a particular place.

    2 Jobs Available
    Actor

    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

    4 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    Multimedia Specialist

    A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

    2 Jobs Available
    Visual Communication Designer

    Individuals who want to opt for a career as a Visual Communication Designer will work in the graphic design and arts industry. Every sector in the modern age is using visuals to connect with people, clients, or customers. This career involves art and technology and candidates who want to pursue their career as visual communication designer has a great scope of career opportunity.

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Advertising Manager

    Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

    2 Jobs Available
    Photographer

    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

    2 Jobs Available
    Social Media Manager

    A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

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    3 Jobs Available
    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    2 Jobs Available
    QA Lead

    A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

    2 Jobs Available
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    IT Consultant

    An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

    2 Jobs Available
    Data Architect

    A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements. 

    2 Jobs Available
    Security Engineer

    The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

    2 Jobs Available
    UX Architect

    A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy. 

     

    2 Jobs Available
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