Question 12 Marks
If $A(-2,2,3)$ and $B(13,-3,13)$ are two points respectively. A point P moves in such a way that $3 PA =2 PB$. Prove that locus of point P is $x^2+y^2$ $+z^2+28 x-12 y+10 z-247=0$.
AnswerLet the coordinates of point P be $(x, y, z)$, then
$\begin{aligned} PA ^2 & =(x+2)^2+(y-2)^2+(z-3)^2 \\ \text { and } PB ^2 & =(x-13)^2+(y+3)^2+(z-13)^2\end{aligned}$
Now according to given condition, $3 PA =2 PB$
$\Rightarrow 9 PA ^2=4 PB^2$
$\begin{aligned} \Rightarrow 9\left\{(x+2)^2+(y-2)^2\right. & \left.+(z-3)^2\right\}=4\left\{(x-13)^2\right. \left.+(y+3)^2+(z-13)^2\right\}\end{aligned}$
$\begin{array}{r}\Rightarrow 9\left[x^2+4 x+4+y^2-4 y+4+z^2-6 z+9\right]= 4\left[x^2-26 x+169+y^2+6 y+9+\right. \left.z^2-26 z+169\right]\end{array}$
$\begin{array}{r}\Rightarrow 9 x^2+36 x+36+9 y^2-36 y+36+9 z^2- 54 z+81 =4 x^2-104 x+676+4 y^2+24 y+36+ 4 z^2-104 z+676\end{array}$
$\begin{array}{r}\Rightarrow 9 x^2+9 y^2+9 z^2+36 x-36 y-54 z+153 =4 x^2+4 y^2+4 z^2-104 x+24 y- 104 z+1388\end{array}$
$\begin{array}{r}\Rightarrow 9 x^2-4 x^2+9 y^2-4 y^2+9 z^2-4 z^2+140 x- 60 y+50 z-1235=0\end{array}$
$\begin{array}{l}\Rightarrow 5 x^2+5 y^2+5 z^2+140 x-60 y+50 z-1235=0 \\ \Rightarrow x^2+y^2+z^2+28 x-12 y+10 z-247=0\end{array}$
Which is required locus of the point P.
View full question & answer→Question 22 Marks
Show that the points $A(0,7,10), B(-1,6,6)$ and $C (-4,9,6)$ are the vertices of an isosceles right - angles triangle.
AnswerLet $A B C$ be a given triangle
$ \begin{aligned}\text { So, }\quad AB^2 & =(-1-0)^2+(6-7)^2+(6-10)^2 \\& =1+1+16=18 \\BC^2 & =(-1+4)^2+(6-9)^2+(6-6)^2 \\& =9+9=0=18 \\CA^2 & =(0+4)^2+(7-9)^2+(10-6)^2 \\& =16+4+16=36 \\\therefore\quad\quad AB & =BC \text { and } AB^2+BC^2=AC^2\end{aligned}$
Hence, we can say that A, B and C be the vertices of an isosceles right - angled triangle.
Hence proved.
View full question & answer→Question 32 Marks
Find the perimeter of a triangle whose verticies have the coordinates $A ( 1 0 , - 2 , 8 ), B ( 8 , 0 , 7 )$ and $C (10,2,8).$
AnswerLet $ABC$ be a triangle and $AB , BC$ and $CA$ be its three edges.
$\begin{aligned}\text { So, } AB & =\sqrt{(10-8)^2+(-2-0)^2+(8-7)^2} \\& =\sqrt{(2)^2+(-2)^2+(1)^2}=\sqrt{4+4+1}=\sqrt{9}=3 \\BC & =\sqrt{(8-10)^2+(0-2)^2+(7-8)^2} \\& =\sqrt{(-2)^2+(-2)^2+(-1)^2}=\sqrt{4+4+1}=\sqrt{9}=3 \\\text { and } CA & =\sqrt{(10-10)^2+(2+2)^2+(8-8)^2} \\& =\sqrt{0^2+4^2+0^2}=\sqrt{16}=4\end{aligned}$
So, Perimeter of $\Delta A B C=A B+B C+C A$
$=3+3+4=10$
View full question & answer→Question 42 Marks
Find the locus of a point such that, the difference of its distances from the points $A(3,4,5)$ and $B (-1,3,-7)$ is $2 \lambda^2.$
AnswerLet the required point be $P (x, y, z)$
So, $PA ^2- PB ^2=2 \lambda^2$
$\begin{aligned} {\left[(x-3)^2+(y-4)^2+\right.} & (z-5)^2-(x+1)^2- \left.(y-3)^2-(z+7)^2\right]=2\lambda^2\end{aligned}$
$\begin{array}{r}\Rightarrow x^2-6 x+9+y^2-8 y+16+z^2-10 z+25- x^2-2 x-1-y^2+6 y-9-z^2 -14 z-49=2 \lambda^2\end{array}$
$\begin{array}{l}-6 x+9-8 y+16-10 z+25-2 x-1+ 6 y-9-14 z-49=2 \lambda^2\end{array}$
$\begin{array}{r}\Rightarrow(-6 x-2 x)+(-8 y+6 y)+(-10 z-14 z)- 9=2 \lambda^2\end{array}$
$\begin{array}{l}\Rightarrow-8 x-2 y-24 z-9=2 \lambda^2 \\ \Rightarrow-8 x-2 y-24 z-9-2 \lambda^2=0 \\ \Rightarrow 8 x+2 y+24 z+2 \lambda^2+9=0\end{array}$
Which is the required locus of the point.
View full question & answer→Question 52 Marks
Find the point on $Z-axis$ which is equidistant from points $A(1,5,7)$ and $B(5,1,-4)$.
AnswerLet $A (1,5,7)$ and $B (5,1,-4)$ and the required point on $Z-\text {axis}$ be $P (0,0, z)$.
$\therefore\quad$ According to question,
$AP = BP$
$\Rightarrow \quad AP ^2= BP ^2$
$\begin{aligned} \Rightarrow(1-0)^2+(5 & -0)^2+(7-z)^2 =(5-0)^2+(1-0)^2+(-4-z)^2\end{aligned}$
$\Rightarrow 1+25+49+z^2-14 z=25+1+16+z^2+8 z$
$\Rightarrow \quad 49-16=8 z+14 z$
$\Rightarrow \quad 33=22 z$
$\Rightarrow \quad z=\frac{33}{22}=\frac{3}{2}$
Hence, required point on $Z-\text {axis}$ is $(0,0, z)$
$=\left(0,0, \frac{3}{2}\right)$
View full question & answer→Question 62 Marks
Find the point in $Y Z$ plane which is equidistant from points $A (a, 0,0), B (0, b, 0)$ and $C (0,0, c).$
AnswerLet the point in YZ plane be $P (0, y, z)$
$\begin{aligned}\therefore\quad AP & =\sqrt{a^2+y^2+z^2} \\ \text { and } BP & =\sqrt{(b-y)^2+z^2} \\ CP & =\sqrt{y^2+(c-z)^2}\end{aligned}$
$\because$ According to question,
$\quad\quad AP=BP=CP$
$\Rightarrow \quad AP ^2= BP ^2= CP ^2$
$\therefore$ From $\quad AP ^2= BP ^2,$
$a^2+y^2+z^2=(b-y)^2+z^2$
$\Rightarrow\quad a^2+y^2+z^2=b^2+y^2-2 b y+z^2$
$\Rightarrow \quad 2 b y=b^2-a^2$
$\Rightarrow \quad y=\frac{b^2-a^2}{2 b}$
Again from $AP ^2= CP ^2$
$a^2+y^2+z^2=y^2+(c-z)^2$
$\Rightarrow \quad a^2+y^2+z^2=y^2+c^2+z^2-2 c z$
$\Rightarrow \quad 2 c z=c^2-a^2$
$\therefore \quad z=\frac{c^2-a^2}{2 c}$
$\begin{aligned} \text { So required point } & =(0, y, z) \\ & =\left(0, \frac{b^2-a^2}{2 b}, \frac{c^2-a^2}{2 c}\right)\end{aligned}$
View full question & answer→Question 72 Marks
Show that point $A (1,2,3), B =(-1,-2,-1),$ $C(2,3,2)$ and $D(4,7,6)$ are the vertices of a parallelogram.
AnswerUsing distance formula,
$\begin{aligned}AB & =\sqrt{\left\{(-1-1)^2+(-2-2)^2+(-1-3)^2\right\}} \\& =\sqrt{4+16+16}=\sqrt{36}=6 \\BC & =\sqrt{\left\{(2+1)^2+(3+2)^2+(2+1)^2\right\}} \\& =\sqrt{9+25+9}=\sqrt{43} \\CD & =\sqrt{\left\{(4-2)^2+(7-3)^2+(6-2)^2\right\}} \\& =\sqrt{4+16+16}=\sqrt{36}=6 \\DA & =\sqrt{\left\{(1-4)^2+(2-7)^2+(3-6)^2\right\}} \\& =\sqrt{9+25+9}=\sqrt{43} \\AC & =\sqrt{(2-1)^2+(3-2)^2+(2-3)^2} \\& =\sqrt{1+1+1}=\sqrt{3} \\BD & =\sqrt{(4+1)^2+(7+2)^2+(6+1)^2} \\& =\sqrt{25+81+49}=\sqrt{155}\end{aligned}$
We see in quadrilateral $ABCD.$
$AB = CD =6$ and $BC = DA =\sqrt{43}$ and $AC \neq BD$
Since the opposite sides of quadrilateral are equal and diagonals are unequal, so quadrilateral ABCD is a parallelogram.
Hence given points are vertices of a parallelogram.
View full question & answer→Question 82 Marks
Find the coordinates of a point which is equidistant from points $\text {O (0,0,0), A (a, 0,0)},$ $\text {B (0,b,0), C (0,0,c}).$
AnswerLet the coordinates of required point be $P ( x , y, z).$
So according to question, $OP = PA = PB = PC$
$\begin{array}{l}\therefore \quad OP^2=PA^2 \\\Rightarrow \quad x^2+y^2+z^2=(x-a)^2+(y-0)^2+(z-0)^2 \\\Rightarrow \quad x=\frac{a}{2} \\\text { Similarly, } \quad OP^2=PB^2 \Rightarrow y=\frac{b}{2} \\\text { and } \quad OP^2=PC^2 \Rightarrow z=\frac{c}{2}\end{array}$
$\therefore$ Coordinates of required point is $\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)$
View full question & answer→Question 92 Marks
Prove that following points are collinear :
(i) $(4,-3,-1),(5,-7,6),(3,1,-8)$
(ii) $(0,7,-7),(1,4,-5),(-1,10,-9)$
Answer(i) Let three given points are $A (4,-3,-1), B (5,-$ $7,6)$ and $C(3,1,-8)$ respectively.
$\begin{aligned}\therefore \quad AB & =\sqrt{(5-4)^2+(-7+3)^2+(6+1)^2} \\& =\sqrt{1+16+49}=\sqrt{66} \\BC & =\sqrt{(3-5)^2+(1+7)^2+(-8-6)^2} \\& =\sqrt{4+64+196}=\sqrt{264}=2 \sqrt{66} \\\text { and } CA & =\sqrt{(3-4)^2+(1+3)^2+(-8+1)^2} \\& =\sqrt{1+16+49}=\sqrt{66}\end{aligned}$
$\because \quad BC = AB + AC \quad \therefore$ Given points are collinear.
(ii) Let the three given points be $(0,7,-7),(1,4,-5)$, $(-1,10,-9)$. Let be points are $A, B, C.$
$\begin{aligned}AB & =\sqrt{(0-1)^2+(7-4)^2+(-5+7)^2} \\& =(1+9+4)=\sqrt{14} \\BC & =\sqrt{(-1-1)^2+(10-4)^2+(-9+5)^2} \\& =\sqrt{4+36+16}=\sqrt{56}=2 \sqrt{14} \\CA & =\sqrt{(0+1)^2+(10-7)^2+(-9+7)^2} \\& =\sqrt{1+9+4}=\sqrt{14}\end{aligned}$
$\because \quad BC = AB + AC \quad \therefore$ Given points are collinear.
View full question & answer→Question 102 Marks
Write the distances of point $(1,2,3)$ from the coordinate axes.
AnswerDistance from $x-axis$
$=\sqrt{(1-1)^2+(2-0)^2+(3-0)^2}$
$=\sqrt{4+9}=\sqrt{13}$
Distance from $y-axis$
$\begin{array}{l}=\sqrt{(1-0)^2+(2-2)^2+(3-0)^2} \\=\sqrt{1+9}=\sqrt{10}\end{array}$
Distance from $z-axis$
$\begin{array}{l}=\sqrt{(1-0)^2+(2-0)^2+(3-3)^2} \\=\sqrt{1+4}=\sqrt{5}\end{array}$
View full question & answer→