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 $$ \sqrt {{{b}^{2} + {c}^{2}} }$$.