Question 1 1 / -0
Solution
$${\textbf{Step -1: Finding distance between points .}}$$
$${\text{Let A(4, -5, 1)}}$$
$${\text{B=(3, -4, 0)}}$$
$${\text{C=(6, -7, 3)}}$$
$${\text{D=(7, -8, 4)}}$$
$${\text{ Now let the quadrilateral be ABCD}}$$
$${\text{Then,}}$$
$$AB=-i+j-k |AB|$$
$$=\sqrt3$$
$$BC= 3i-3j+3k |BC|$$
$$=3\sqrt 3$$
$$CD=i-j+k|CD|$$
$$=\sqrt3$$
$$AD=3i-3j+3k|AD|$$
$$=3\sqrt3$$
$$\textbf{Step -2: Checking distances between opposite sides . }$$
$${\text{Hence opposite sides are equal and parallel. Therefore the above}}$$
$${\text{points form a parallelogram. Now all the sides are not equal. Hence}}$$
$${\text{it cannot qualify as a rhombus or square.}}$$
$${\text{Now dot product of AB and BC is not zero. Hence adjacent sides are not perpendicular to each other.}}$$
$${\text{Therefore it is not a rectangle also.}}$$
$${\textbf{ Hence , option B is correct.}}$$
Question 2 1 / -0
Solution
the centroid of the coordinates is
$$\left(\dfrac {x_1+x_2+x_3+x_4}{4},\dfrac{y_1+y_2+y_3+y_4}{4},\dfrac{z_1+z_2+z_3+z_4}{4}\right)$$
Thus by substituting the vertices we get
$$=\left(\dfrac{0+4+0+0}{4},\dfrac{0+0-8+0}{4},\dfrac{0+0+0+12}{4}\right)$$
$$=\left(\dfrac{4}{4},\dfrac{-8}{4},\dfrac{12}{4}\right)$$
$$\therefore$$ the centroid of the coordinates is $$(1,-2,3)$$
Question 3 1 / -0
Solution
In Geometry , we define a point as a location and no size. A line is defined as something that extends infinitely in either direction but has no width and is one dimensional while a plane extends infinitely in two dimensions.
Question 4 1 / -0
Solution
The plane forming the parallelipiped are
$$x=-1,x=1;y=2,y=-1$$ and $$z=5,z=-1$$
Hence, the lengths of the edges of the parallelopiped are
$$1-\left( -1 \right) =2,\left| -1-2 \right| =3$$ and $$\left| -1-5 \right| =6$$
$$($$ length of an edge of the parallelopiped is the distance between the parallel plane sperpendicular to the edge$$)$$
$$\therefore$$ Length of diagonal of the parallelopiped
$$=\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 49 } =7$$
Question 5 1 / -0
Solution
Ans: B
Question 6 1 / -0
Solution
(i) Distance between $$4i + 3j - 6k$$ and $$-2i + j - k$$ is given by $$\sqrt{(4 + 2)^2 + (3 - 1)^2 + (-6 + 1)^2} = \sqrt{36 + 4 + 25} = \sqrt{65}$$
$$\Rightarrow k = 65$$
(ii) Distance between $$-2i + 3j + 5k$$ and $$7i - k$$ would be $$\sqrt{(-2 - 7)^2 + (3)^2 + (5 + 1)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$$
$$\Rightarrow m = 3, n = 14$$
$$\therefore k - (m*n) = 65 - (3 \times 14) = 65 - 42 = 23$$
Question 7 1 / -0
Solution
Let $$P$$ be the required point $$\displaystyle \left ( x,y,z \right )$$ and the point
$$A, B, C$$ and $$O$$ are $$\displaystyle \left ( a,0,0 \right ),\left ( 0,b,0 \right ),\left ( 0,0,c \right )$$ and $$\displaystyle \left ( 0,0,0 \right )$$ ;
We are given that $$\displaystyle PO=PA=PB=PC.$$
Taking $$\displaystyle PO=PA$$ or $$\displaystyle PO^{2}=PA^{2},$$ we get
$$\displaystyle x^{2}+y^{2}+z^{2}=\left ( x-a \right )^{2}+y^{2}+z^{2}$$
$$\displaystyle 0=a^{2}-2ax$$ i.e. $$\displaystyle x=\dfrac {a}{2}$$
Similarly taking $$\displaystyle PO^{2}=PB^{2}$$ and $$\displaystyle PO^{2}=PC^{2},$$ we get
$$\displaystyle y=\dfrac {b}{2}$$ and $$\displaystyle z=\dfrac {c}{2}$$
Question 8 1 / -0
Solution
The vertices of tetrahedron are $$(0,0,0) , (\alpha , 5,6) , (1,\beta , 4) , (3,2, \gamma)$$
The centroid of tetrahedron is $$(\frac{\alpha+4}{4},\frac{\beta+7}{4},\frac{\gamma+10}{4})$$
Given that centroid is $$1,-1,2$$
By equating , we get $$ \alpha=0 , \beta=-11 , \gamma=-2$$
Since $$\alpha =0$$ , $${ \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 }={ \beta }^{ 2 }+{ \gamma }^{ 2 }$$
Therefore the correct option is $$B$$
Question 9 1 / -0
Solution
Let $$\lambda$$ be the ratio in which yz-plane divides the line joining the points $$(-2,4,7)$$ and $$(3,-5,8)$$
The co-ordinates of any point on the line joining the two points are
$$\displaystyle \left ( \frac{3\lambda -2}{\lambda +1},\frac{-5\lambda +4}{\lambda +1},\frac{8\lambda +7}{\lambda +1} \right )$$
If the point is in the $$yz$$ - plane, then its $$x$$-coordinate should be zero.
$$\therefore \displaystyle \frac{3\lambda -2}{\lambda +1}=0$$ or $$\displaystyle 3\lambda -2=0 $$
$$ \therefore \lambda =\dfrac{2}{3}.$$
Question 10 1 / -0
Solution
Vertices of triangle are $$(0,4,0),(3,4,0)$$ and $$(0,4,4)$$ then perimeter=?
here AB=$$\sqrt{(3-0)^2+(4-4)^2+(0-0)^2}$$
$$=3$$
BC$$=\sqrt{(3-0)^2+(4-4)^2+(0-4)^2}$$
$$=\sqrt{9+16}$$
$$=5$$
CA$$=\sqrt{(0-0)^2+(4-4)^2+(0-4)^2}=4$$
used distance formula b/w two points
$$(x_1,y_1,z_1) and (x_2,y_2,z_2)$$
$$=\sqrt{(x_2-x_2)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$
$$ perimeter =ABC+BC+CA$$
$$=3+5+4$$
$$=12\ units$$
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