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Introduction to Three Dimensional Geometry Test - 35

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Introduction to Three Dimensional Geometry Test - 35
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  • Question 1
    1 / -0
    The foot of the perpendicular from the point $$A(7, 14, 5)$$ to the plane $$2x+4y-z=2$$ is?
    Solution
    Let N be the foot of the perpendicular drawn from the point $$A(7, 14, 5)$$ and perpendicular to the plane $$2x+4y-z=2$$

    Then, the equation of the line PN is $$\dfrac{x-7}{2}=\dfrac{y-14}{4}=\dfrac{z-5}{-1}=\lambda$$ (say)

    Let the coordinates of N be $$N(2\lambda +7, 4\lambda +14, -\lambda +5)$$

    Since N lies on the plane $$2x+4y-z=2$$, so

    $$2(2\lambda +7)+4(4\lambda +14)-(-\lambda +5)=2$$

    $$\Rightarrow 21\lambda =-63$$

    $$\Rightarrow \lambda =-3$$

    $$\therefore$$ required foot of the perpendicular is

    $$N(-6+7, -12+14, 3+5)$$, i.e., $$N(1, 2, 8)$$.
  • Question 2
    1 / -0
    The coordinates of the point where the line through the points $$A(5,1,6)$$ and $$B(3,4,1)$$ crosses the yz-plane is
    Solution
    Equation of the line passing through the points $$A(5, 1, 6)$$ and $$B(3,4, 1)$$ is

    $$\dfrac{x-5}{2}=\dfrac{4-1}{-3} = \dfrac{7-6}{5}$$ ....(i)

    Let the line (i) cross the yz plane at point $$(o, h, k)$$

    Then $$(o, h, k)$$ point lie on the line (I)

    So, $$\dfrac{0-5}{2} = \dfrac{h-1}{-3} = \dfrac{k-6}{5}$$

    Then, $$\dfrac{k-6}{5} = - \dfrac{5}{2}$$

    or, $$k-6 = -\dfrac{25}{2}$$

    or, $$k = 6- \dfrac{-25}{2} =\dfrac{12-25}{2} = -\dfrac{13}{2}$$

    Again, $$\dfrac{h-1}{-3} = \dfrac{-5}{2}$$

    or, $$h-1 = \dfrac{15}{2}$$

    or, $$h=\dfrac{15}{2} + 1 = \dfrac{17}{2}$$

    Therefore the crossing points is $$\left(0, \dfrac{17}{2}, -\dfrac{13}{2}\right)$$
  • Question 3
    1 / -0
    The point on the line $$\dfrac {x - 2}{1} = \dfrac {y + 3}{-2} = \dfrac {z + 5}{-2}$$ at a distance of 6 from the point $$\left ( 2,-3,-5 \right )$$ is
    Solution

  • Question 4
    1 / -0
    The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are $$d_1,d_2,d_3$$. Then $$d_1^2 +d_2^2 + d_3^2 = kd^2$$ where K is
    Solution

  • Question 5
    1 / -0
    The shortest distance between the lines $$\dfrac {x - 3}{2} = \dfrac {y + 15}{-7} = \dfrac {z - 9}{5}$$ and $$\dfrac {x + 1}{2} = \dfrac {y - 1}{1} = \dfrac {z - 9}{-3}$$ is
    Solution

  • Question 6
    1 / -0
    The ratio in which the plane $$\overrightarrow{r} \cdot \left (\overrightarrow{i}  - 2 \overrightarrow{j}  + 3 \overrightarrow{k}   \right ) = 17$$ divides the line joining the points $$ -2 \overrightarrow{i}  + 4 \overrightarrow{j}  + 7 \overrightarrow{k} $$ and  $$ 3 \overrightarrow{i}  - 5 \overrightarrow{j}  + 8 \overrightarrow{k} $$
    Solution
    Let the plane $$\overrightarrow{r} \cdot \left ( \overrightarrow{i}  - 2 \overrightarrow{j}  + 3 \overrightarrow{k} \right ) = 17 $$ divided the line joining the points  $$ -2 \overrightarrow{i}  + 4 \overrightarrow{j}  + 7 \overrightarrow{k} $$ and  $$ 3 \overrightarrow{i}  - 5 \overrightarrow{j}  + 8 \overrightarrow{k} $$ in the ratio $$t : 1$$ at point P.
    Therefore, point P is
    $$\dfrac{3t - 2} {t + 1} \overrightarrow{i} +  \dfrac{-5t + 4} {t + 1} \overrightarrow{j} + \dfrac{8t + 7} {t + 1} \overrightarrow{k}$$
    This lies on the given plane
    $$\therefore \dfrac{3t - 2} {t + 1} (1) +  \dfrac{-5t + 4} {t + 1} (- 2) + \dfrac{8t + 7} {t + 1} (3) = 17$$
    Solving, we get
    $$t = \dfrac{3} {10}$$
  • Question 7
    1 / -0
    The point on the line $$\frac{x - 2} {1} = \frac{y + 3} {-2} = \frac{z + 5} {-2} $$ at a distance of 6 from the point $$\left ( 2, -3, -5 \right )$$ is
    Solution
    b. Direction cosines of the given line are $$\frac{1} {3},  - \frac{2} {3}, -\frac{2} {3}$$
    Hence, the equation of line can be point in the form $$\frac{x - 2} {1/3} =   \frac{y + 3} {-2/3} = -\frac{z + 5} {-2/3} = r $$
    Therefore, any point on the line is $$\left (2 + \frac{r} {3}, - 3 - \frac{2r} {3}, - 5 - \frac{2r} {3}  \right )$$, where $$r = \pm 6$$.
    Points are$$\left (4, -7, -9 \right )$$ and $$\left (0, 1, -1  \right )$$
  • Question 8
    1 / -0
    The points $$A(1,2,-1),B(2,5,-2),C(4,4,-3)$$ and $$D(3,1,-2)$$ are
    Solution
    AB$${=}$$ $$\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$$
    AB$${=}$$ $$\sqrt{{(1)}^{2}+{(3)}^{2}+{(-1)}^{2}}$$
    AB$${=}$$ $$\sqrt{11}$$
    Similarly you find that BC$${=}$$ $$\sqrt{6}$$  CD$${=}$$ $$\sqrt{11}$$  and DA$${=}$$$$\sqrt{6}$$
    Hence opposite sides of quadrilateral are equal, Now we check the diagonals
    AC$${=}$$ $$\sqrt{{(4-1)}^{2}+{(4-2)}^{2}+{(-3+1)}^{2}}$$
    AC$${=}$$ $$\sqrt{17}$$
    similarly BD$${=}$$ $$\sqrt{17}$$ 
    Diagonals are not equal
    direction ratio of line passing through AB is (1,3,-1)
    direction ratio of line passing through  BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
    opposite sides are equal and diagonal are equal
    hence it is rectangle
  • Question 9
    1 / -0
    Let $$A \left ( 2\hat{i}+3\hat{j}+5\hat{k} \right )B\left ( -\hat{i}+3\hat{j}+2\hat{k} \right ) $$and $$ C \left ( \lambda \hat{i}+5\hat{j}+\mu\hat{k} \right )$$ are vertices of a triangle and its median through $$A$$ is equally inclined to the positive directions of the axes. The value of $$\lambda+\mu$$ is equal to
    Solution
    We have, 
    $$ \vec{AD} = \dfrac 12 \left( \vec b + \vec c - 2 \vec a \right) $$

    $$ \vec{AD} = \dfrac{ \lambda -5 }{2} \hat i + \hat j + \dfrac{\mu-8}{2} \hat{k} $$ 
    The line is equally inclined to the axes. Hence, its direction ratios must be of the form $$1:1:1$$.
    Hence, $$\lambda= 7 $$ and $$\mu = 10 $$

  • Question 10
    1 / -0
    If $$(0, b, 0)$$ is the centroid of the triangle formed by the points $$(4, 2, -3)$$ , $$({a}, -5, 1)$$ and $$(2, -6, 2)$$ . If $$a ,b$$ are the roots of the quadratic equation $$ x^2+px+q = 0 $$, then $$p,q$$ are 
    Solution
    Since $$a,b$$ are the roots of the equation
    $$x^{2}+px+q=0$$
    $$\Rightarrow a+b=-p$$ and $$ab=q$$
    Centroid of triangle is $$\displaystyle \left(\dfrac{a+6}{3},-3,0\right)$$
    Given, centroid $$(0,b,0)$$
    Comparing, we get $$  b=-3$$
    and $$\dfrac{a+6}{3}=0$$
    $$\Rightarrow a=-6$$
    Hence, $$p=9, q=18$$

    Hence, option A.
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