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Introduction to Three Dimensional Geometry Test - 15

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Introduction to Three Dimensional Geometry Test - 15
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  • Question 1
    1 / -0
    If $$(1, 1, a)$$ is the centroid of the triangle formed by the points $$(1, 2, -3)$$ , $$(\mathrm{b}, 0, 1)$$ and $$(-1, 1, -4)$$ then $$a-b$$ $$=$$
    Solution

    The coordinates of the vertices of the triangle are given by $$(1,2,-3) , (b,0,1), (-1,1,-4)$$

    Accordingly the coordinates of the centroid of this triangle will be given by 

    ($$ \dfrac{b}{3}, 1 , -2 $$)

    Hence, $$ \dfrac{b}{3} $$ $$= 1$$ Or, $$b= 3$$

    and $$a = -2$$

    So $$a- b = -5$$

  • Question 2
    1 / -0
    If the extremities of a diagonal of a square are $$(1, -2, 3)$$ and $$(2, -3, 5)$$, then area of the square is
    Solution
    Let the extremities of the diagonal of a square be $$(1,-2,3)$$ and $$B(2,-3,5)$$.
    Then diagonal $$AB$$ is given by $$ {({1}^{2} + {1}^{2} + {2}^{2})}^{0.5} $$ = $$ \sqrt{6} $$
    Hence, length of the side $$ =\text{length of diagonal}\div\sqrt2= \sqrt {3} $$
    So, area of square will be $$ \sqrt{3} \times \sqrt{3}  = 3$$
  • Question 3
    1 / -0
    If $$A= (1, 2, 3), B = (2, 3, 4)$$ and $$AB$$ is produced upto $$C$$ such that $$2AB = BC$$, then $$C =$$
    Solution
    Let the point $$C$$ be $$(i,j,k)$$.
    Since, $$B$$ divides $$AC$$ in the ratio $$1:2$$
    Coordinates of $$B$$ should be $$\left ( \dfrac{2+i}{3} , \dfrac{4+j}{3} , \dfrac{k+6}{3}\right) $$
    Comparing the values given already for $$B$$, we get, 
    $$i = 4$$, $$j = 5$$ and $$k = 6$$
  • Question 4
    1 / -0
    If the points $$A(3, -2, 4)$$, $$B(1, 1, 1)$$ and $$C(-1, 4, -2)$$ are collinear, then the ratio in which $$C$$ divides $$AB$$ is 
    Solution
    Let $$C$$ divide $$AB$$ in the ratio $$x:y$$. 
    Let us compare the $$x$$-coordinate of $$C$$ by using section formula: 
    $$-1 = \dfrac {x\times 1 + y\times3}{x+y} $$
    $$ \Rightarrow x+3y=-x-y $$
    $$\Rightarrow x = -2y $$
    Hence, point $$C$$ divides $$AB$$ in the ratio $$-2 :1 $$. As the ratio is negative, it means $$C$$ divides $$AB$$ externally
  • Question 5
    1 / -0
    If $$A = (2, -3, 1), B = (3, -4, 6)$$ and $$C$$ is a point of trisection of $$AB$$, then $${C}_{{y}}=$$
    Solution
    Given, $$C$$ is a point of trisection of $$AB$$.
    Hence, $$C$$ either divides $$AB$$ in the ratio $$2:1$$ or $$1:2$$.
    Case 1: $$C$$ divides in the ratio $$2:1$$.
    The coordinates of $$C$$ will be $$\left(\dfrac{8}{3}, -\dfrac{11}{3}, \dfrac{13}{3}\right)$$
    Case 2: $$C$$ divides in the ratio $$1:2$$.
    The coordinates of $$C$$ will be $$\left(\dfrac{7}{3}, -\dfrac{10}{3}, \dfrac{8}{3}\right)$$

    Hence, either $$C_y =  -\dfrac{11}{3}$$ or $$-\dfrac{10}{3}$$
  • Question 6
    1 / -0
    Two opposite vertices of a square are $$(2, -3, 4)$$ and $$(4, 1, -2)$$. The length of the side of the square is
    Solution
    Length of diagonal $$=$$ distance between the opposite vertices of the square $$\\=\sqrt{(2-4)^2+(-3-1)^2+(4+2)^2}=\sqrt{56}=d$$ (say)
    Hence, length of sides of square $$=\dfrac{d}{\sqrt 2}=\sqrt{28}=2\sqrt 7$$
  • Question 7
    1 / -0
    The point $$P$$ is on the $$y$$-axis. If $$P$$ is equidistant from $$(1,2, 3)$$ and $$(2,3, 4)$$, then $$P_{y}=$$
    Solution
    Let the point be $$(0,y,0)$$
    Since, it is equidistant from the $$2$$ points.
    $$\Rightarrow  (1+ (y-2)^{2} + 9) = (4 + (y-3)^{2} + 16) $$
    $$\Rightarrow (10+y^2-4y+4)= (20+y^2-6y+9)$$
    $$\Rightarrow y^2-4y+14=y^2-6y+29$$
    $$\Rightarrow 2y=15$$
    $$\Rightarrow y =$$ $$ \displaystyle \dfrac{15}{2} $$
  • Question 8
    1 / -0
    $$A = (1, -2, 3)$$ , $$B =$$ (2, 1, 3), $$C =$$ (4, 2, 1) and $$G= (-1,3, 5)$$ is the centroid of the tetrahedron $$ABCD$$. Then the fourth coordinate is
    Solution
    Given : $$A=(1,-2,3), B=(2,1,3), C=(4,2,1)$$ and $$G=(-1,3,5)$$
    Let $$D=(a,b,c)$$
    $$G$$ is centroid of the tetrahedron $$ABCD$$ then,
    $$G =\dfrac{A+B+C+D}{4}$$
    $$\implies G=\dfrac{(1,-2,3)+(2,1,3)+(4,2,1)+(a,b,c)}{4}$$
    $$\implies 4G=(1+2+4+a,-2+1+2+b,3+3+1+c)$$
    $$\implies (-4, 12, 20)=(7+a, 1+b, 7+c)$$
    $$\implies 7+a=-4, 1+b=12, 7+c=20$$
    $$\implies a=-11, b=11, c=13$$
    Hence, $$D=(-11,11,13)$$ is the fourth coordinate.
  • Question 9
    1 / -0
    $$XOZ$$ plane divides the join of $$(2,3,1)$$ and $$(6,7,1)$$ in the ratio
    Solution
    Let the plane divide the line in the ratio $$p : 1$$.
    A point that divides the line joining these $$2$$ points in the ratio $$p : 1$$ is given by, $$\left ( \dfrac{6p+2}{p+1} , \dfrac{7p+3}{p+1}, \dfrac{p+1}{p+1} \right)$$
    Since, this point has to lie on the $$zx$$ plane, so 
    $$7p +3 =0$$
    $$\dfrac{p}{1} =$$ $$ \dfrac{-3}{7} $$
  • Question 10
    1 / -0
    The shortest distance of $$(a,b,c)$$ from $$x$$-axis is
    Solution
    The shortest distance of a point is always the perpendicular distance.
    The projection of the point $$(a,b,c)$$ on the $$x$$-axis is $$(a,0,0)$$.
    This distance is given by $$ \sqrt {{{b}^{2} + {c}^{2}} }$$.
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