Self Studies
Self Studies
Self Studies
Self Studies

Introduction to Three Dimensional Geometry Test - 29

Result Self Studies

Introduction to Three Dimensional Geometry Test - 29
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    A plane meets the co-ordinate axes in A,B,C such that the centroid of the triangle ABC is the point $$(p,q,r)$$. The equation of the plane is 
    Solution

    The equation of plane is

    $$\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{z}{C}=1$$           …….   (1)

    Using centroid formula in XYZ plane is

    $$\left( x=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},y=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},z=\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$

    Then,

    $$\left( \dfrac{A}{3},\dfrac{B}{3},\dfrac{C}{3} \right)=\left( p,q,r \right)$$

    Therefore,

    $$A=3p,\,\,\,B=3q,\,\,C=3r$$

    Put the value of $$A,B$$ and $$C$$ in equation of plane (1) and we get

    $$\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{z}{C}=1$$

    $$\dfrac{x}{3p}+\dfrac{y}{3q}+\dfrac{z}{3r}=1$$

    Hence,

    $$\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=3$$

    option (D) is correct answer.

     

  • Question 2
    1 / -0
    If a point $$P$$ from where line drawn cuts coordinates axes at $$A$$ and $$B$$ (with $$A$$ on $$x-$$axis and $$B$$ on $$y-$$axis ) satisfies $$\alpha \dfrac{x^{2}}{PB^{2}}+\beta \dfrac{y^{2}}{PA^{2}}=1$$, then $$\alpha+\beta$$ is
    Solution
    $$\alpha\dfrac{x^2}{PB^2}+\beta\dfrac{y^2}{PA^2}=1$$
    $$PA=\sqrt{4+4}=2\sqrt 2$$
    $$PB=2\sqrt 2$$
    put value of $$PA$$ & $$PB$$ in given eqn
    $$\alpha\dfrac{4}{8}+\beta\dfrac{4}{8}=1$$
    $$\alpha+\beta=2$$ Option B
    You can take any value of point $$A$$ & $$B$$ 
    Answer will remain same.

  • Question 3
    1 / -0
    The distance between the points $$(\cos \theta, \sin \theta)$$ and $$(\sin \theta - \cos \theta)$$ is
    Solution
    Distance between the point $$(\cos\theta, \sin\theta)$$ and $$(\sin\theta, -\cos\theta)$$ is :
    $$ = \sqrt{(\sin\theta - \cos\theta)^2 + (-\cos\theta - \sin\theta)^2}$$
    $$ = \sqrt{1 - 2 \sin\theta \cos\theta + 1 + 2 \sin\theta \cos\theta}$$
    $$(\cos^2\theta + \sin^2\theta = 1)$$
    $$= \sqrt{2}$$
    Option B is the correct answer
  • Question 4
    1 / -0
    The point equidistant from the point $$O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$$ and $$C(0, 0, c)$$ has the coordinates
    Solution
    $$P(x, y, z)$$
    $$PO=\sqrt{x^2+y^2+z^2}$$
    $$PA=\sqrt{(a-x)^2+(-y)^2+(-z)^2}=\sqrt{(a-z)^2+y^2+z^2}$$
    $$PB=\sqrt{x^2+(b-y)^2+z^2}$$
    $$PC=\sqrt{x^2+y^2+(C-z)^2}$$
    $$\Rightarrow PO=PA$$
    $$\sqrt{x^2+y^2+z^2}=\sqrt{(a-x)^2+y^2+z^2}$$
    $$x^2+y^2+z^2=(a-x)^2+y^2+z^2$$
    $$x^2=(a-x)^2$$
    $$x=a-x$$
    $$\Rightarrow x=\dfrac{a}{2}$$
    $$y=b/2$$
    $$z=c/2$$.
  • Question 5
    1 / -0
    $$A$$ point $$C$$ with position vector $$\frac{{3a + 4b - 5c}}{3}$$ (where a,b and c are non co-planar vectors) divides the line joining $$A$$ and $$B$$ in the ratio $$2:1$$. If the position vector of $$A$$ is $$a-2b+3c$$, then the position vector of $$B$$ is
    Solution
    $$\begin{array}{l} a-2b+3c \\ \frac { { 3a+4b-5c } }{ 3 }  \\ \vec { c } =\frac { { 2\vec { b } +\vec { a }  } }{ 3 }  \\ \vec { b } =\frac { { 3\vec { c } -\vec { a }  } }{ 2 }  \\ =\frac { { \left( { 3\vec { a } +4\vec { b } -5\vec { c }  } \right) -\left( { \vec { a } -2\vec { b } +3\vec { c }  } \right)  } }{ 2 }  \\ =\vec { a } +3\vec { b } -4\vec { c }  \end{array}$$
  • Question 6
    1 / -0
    If the distance between a point P and the point (1, 1, 1) on the line $$\frac{{x\, - \,1}}{3}\, = \,\frac{{y - \,1}}{4}\, = \,\frac{{z\, - 1}}{{12}}$$ is 13, then the coordinates of P are
    Solution
    $$\begin{array}{l} According\, to\, question: \\ we\, have\, the\, line\, equ=\frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } \, and\, point\, (P)\, is\, 13. \\ Now, \\ line\, of\, equ: \\ \Rightarrow \frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } =13 \\ Now\, find\, x,y\, \& \, z\, \, coordinates: \\ \, \Rightarrow \frac { { x-1 } }{ 3 } =13 \\ \, \, \, \, \, \, \, \therefore \, \, \, \, x=39+1=40 \\ \Rightarrow \frac { { y-1 } }{ 4 } =13 \\ \, \, \, \, \, \, \, \therefore \, \, y=\, 52+1=53 \\ \Rightarrow \frac { { z-1 } }{ { 12 } } =13 \\ \, \, \, \, \, \, \, \, \therefore \, \, z=156+1=157 \\ Now,\, we\, get\, \, new\, coordinates\, of\, P:\, \, \, \left( { 40,53,157 } \right) \, \,  \\ so\, that\, \, the\, correct\, option\, is\, D. \end{array}$$
  • Question 7
    1 / -0
    The xy-plane divides the line joining the points (-1, 3, 4) and (2,-5,6).
    Solution
    The ratio that xy-plane divides the line joining the points $$(x_1, y_1, z_1)$$ and $$(x_2, y_2, z_2) = -z_1 : z_2$$

    IF the result is positive, it divides internally otherwise externally

    The ratio that xy-plane divides the line joining the points $$(-1, 3, 4)$$ and $$(2, -5, 6) = -4 : 6 = -2 : 3$$

    i.e., It divides externally in the ratio $$2:3$$
  • Question 8
    1 / -0
    Let $$O$$ be the origin and $$P$$ be the point at a distance $$3$$ units from origin. If d.x.s' of OP are 1, - 2, - 2, then coordinates of P is given by 
    Solution
    $$r=3$$,directions $$(l,m,n)=(\cfrac { 1 }{ 3 } ,\cfrac { -2 }{ 3 } ,\cfrac { -2 }{ 3 } )$$
    $$\therefore$$ $$P(x,y,z)$$ be coordinates of point .
    $$therefore (x-0)=\cfrac { 1 }{ 3 } \times 3,\quad y-0=\cfrac { -2 }{ 3 } \times 3,\quad z-0=\cfrac { -2 }{ 3 } \times 3\\ (x,y,z)=(1,-2,-2)$$
  • Question 9
    1 / -0

    Directions For Questions

    In a parallelogram OABC, vectors $$\vec{a}, \vec{b}, \vec{c}$$ are respectively the position vectors of vertices A, B, C with reference to O as origin. A point E is taken on the side BC which divides it in the ratio of $$2:1$$ internally. Also, the line segment AE intersect the line bisecting the angle O internally in point P. If CP, when extended meets AB in point F. Then?

    ...view full instructions

    The vector $$\vec{AF}$$, is given by?
    Solution

  • Question 10
    1 / -0
    The position vector of the vertices of a triangle $$ABC$$ are $$\hat { i } ,\hat { j } ,\hat { k } $$ then the position vector of its orthocentre is
    Solution
    We have, the position vectors of the vertices of a triangle $$ABC $$are $$\hat { i } ,\hat { j } ,\hat { k } $$.

    Let $$O$$ be the fixed point.
    $$\hat { i }$$ = position vector of $$A=\overrightarrow{OA}$$
    $$\hat { j }$$ = position vector of $$B=\overrightarrow{OB}$$
    $$\hat { k }$$ = position vector of $$C=\overrightarrow{OC}$$

    Let $$AD$$ be the median ofthe triangle $$ABC$$.
    Since, $$D$$ is the mis point of $$BC$$.
    Position vector of $$D=$$ vector  $$OD=\dfrac{(\overrightarrow{OB}+\overrightarrow{OC})}{2}$$

    Now, position vector of $$G$$,
    $$=\dfrac{2(\overrightarrow{OD})+1(\overrightarrow{OA})}{3}$$
    $$=\dfrac{2(\dfrac{\overrightarrow{OB}+\overrightarrow{OC}}{2})+\overrightarrow{OA}}{3}$$
    $$=\dfrac{{\overrightarrow{OA}+\overrightarrow{OB}}+\overrightarrow{OC}}{3}$$

    So,
    $$=\dfrac{{\hat{i}+\hat{j}}+\hat{k}}{3}$$

    Hence, the position vector of the orthocentre is $$\dfrac{1}{3}\left(\hat i+\hat j + \hat k\right)$$.

Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now