3 Marks Question

Question 13 Marks

If $A$ and $B$ be the points $(3, 4, 5)$ and $(-1, 3, -7),$ respectively, find the equation of the set of points $P$ such that $PA^2 + PB^2 = k^2,$ where $k$ is a constant.

Answer

The equation of the set of points $P$ such that $PA^2 + PB^2 = k^2,$ where $k$ is a constant
Given: The points $A (3, 4, 5)$ and $B (-1, 3, -7)$
$\Rightarrow x_1 = 3, y_1 = 4, z_1 = 5; x_2 = -1, y_2 = 3, z_2 = -7;$
$PA^2 + PB^2 = k^2 …(i)$
Let the point be $P (x, y, z).$
Now, by Distance Formula, we know that the distance between two points $P (x_1, y_1, z_1)$ and $Q (x_2, y_2, z_2)$ is given by $\mathrm{PQ}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$.
So, $PA =\sqrt{(3-\mathrm{x})^{2}+(4-\mathrm{y})^{2}+(5-\mathrm{z})^{2}}$
And $PB =\sqrt{(-1-\mathrm{x})^{2}+(3-\mathrm{y})^{2}+(-7-\mathrm{z})^{2}}$
Now, substituting these values in (i), we have
$[(3 - x)^2 + (4 - y)^2 + (5 - z)^2] + [(-1 - x)^2 + (3 - y)^2 + (-7 - z)^2] = k^2$
$\Rightarrow [(9 + x^2 – 6x) + (16 + y^2 – 8y) + (25 + z^2 – 10z)] + [(1 + x^2 + 2x) + (9 + y^2 – 6y) + (49 + z^2 + 14z)] = k^2$
$\Rightarrow 9 + x^2 – 6x + 16 + y^2 – 8y + 25 + z^2 – 10z + 1 + x^2 + 2x + 9 + y^2 – 6y + 49 + z^2 + 14z = k^2$
$\Rightarrow 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = k^2$
$\Rightarrow 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = k^2 – 109$
$\Rightarrow 2 (x^2 + y^2 + z^2 – 2x – 7y + 2z) = k^2 – 109$
$\Rightarrow x^2 + y^2 + z^2 - 2x - 7y + 2z =\frac{k^{2}-109}{2}$

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Question 23 Marks

Find the length of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

Answer

ABC is a triangle with vertices A (0, 0, 6), (0, 4, 0) and C (6, 0, 0).
Let points D, E and F are the mid-points of BC, AC and AB, respectively. So, AD, BE and CF will be the medians of the triangle.

Coordinates of point $D = \left( \frac { 0 + 6 } { 2 } , \frac { 4 + 0 } { 2 } , \frac { 0 + 0 } { 2 } \right)$ = (3, 2, 0)
$\left[ { \because \text { coordinates of mid-point } } { \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } , \frac { y _ { 1 } + y _ { 2 } } { 2 } , \frac { z _ { 1 } + z _ { 2 } } { 2 } \right) } \right]$
Coordinates of point $E = \left( \frac { 0 + 6 } { 2 } , \frac { 0 + 0 } { 2 } , \frac { 6 + 0 } { 2 } \right)$ = (3, 0, 3)
and coordinates of point$F = \left( \frac { 0 + 0 } { 2 } , \frac { 0 + 4 } { 2 } , \frac { 6 + 0 } { 2 } \right) = ( 0,2,3 )$
Now, length of median
AD = Distance between point A and D
$A D = \sqrt { ( 0 - 3 ) ^ { 2 } + ( 0 - 2 ) ^ { 2 } + ( 6 - 0 ) ^ { 2 } }$
[$\because$ distance $= \sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } } ]$
$= \sqrt { 9 + 4 + 36 }$
$= \sqrt { 49 } = 7$
Similarly, $B E = \sqrt { ( 0 - 3 ) ^ { 2 } + ( 4 - 0 ) ^ { 2 } + ( 0 - 3 ) ^ { 2 } }$
$= \sqrt { 9 + 16 + 9 } = \sqrt { 34 }$
and $C F = \sqrt { ( 6 - 0 ) ^ { 2 } + ( 0 - 2 ) ^ { 2 } + ( 0 - 3 ) ^ { 2 } }$
$= \sqrt { 36 + 4 + 9 } = \sqrt { 49 } = 7$
Hence, length of the medians are 7, $\sqrt { 34 }$ and 7.

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Question 33 Marks

Find the equation of the set of the point $P$, the sum of whose distance from $A(4, 0, 0)$ and $B(-4, 0, 0)$ is equal to $10.$

Answer

Let a point $P(x, y, z)$ such that $PA + PB = 10$

$\Rightarrow \sqrt { ( x - 4 ) ^ { 2 } + ( y - 0 ) ^ { 2 } + ( z - 0 ) ^ { 2 } }$$+ \sqrt { ( x + 4 ) ^ { 2 } + ( y - 0 ) ^ { 2 } + ( z - 0 ) ^ { 2 } } = 10$
[$\because$ distance $= \sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } } ]$
$\Rightarrow \sqrt { x ^ { 2 } - 8 x + 16 + y ^ { 2 } + z ^ { 2 } }$$+ \sqrt { x ^ { 2 } + 8 x + 16 + y ^ { 2 } + z ^ { 2 } } = 10$
$\Rightarrow \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 8 x + 16 }$$= 10 - \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 8 x + 16 }$
On squaring sides, we get
$x^2 + y^2 + z^2 - 8x +16 = 100 + x^2 + y^2 + z^2 + 8x +16$
$- 20 \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 8 x + 16 }$
$\Rightarrow - 16 x - 100 = - 20 \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 8 x + 16 }$
$\Rightarrow \quad 4 x + 25 = 5 \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 8 x + 16 }$ [dividing both sides by -4]
Again squaring on both sides, we get
$16x^2 + 200x + 625 = 25(x^2 + y^2 + z^2 + 8x + 16)$
$\Rightarrow 16x^2 + 200x + 625 = 25x^2 + 25y^2 + 25z^2 + 200x + 400$
$\Rightarrow 9x^2 + 25y^2 + 25z^2 - 225 =0$

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Question 43 Marks

Find the equation of the set of points $P$ such that $PA^2 + PB^2 =2k^2,$ where $A$ and $B$ are the points $(3, 4, 5) $and $(-1, 3, -7),$ respectively.

Answer

Given points are $A(3, 4, 5)$ and $B(-1, 3, -7).$
Let the coordinates of point $P$ be $(x, y, z).$
Here, $PA^2 = (3 - x)^2 + (4 - y)^2 + (5 - z)^2$
[$\because$ distance between two points$= \sqrt { \left( x _ { 2 } - x _ { 1 } \right) ^ { 2 } + \left( y _ { 2 } - y _ { 1 } \right) ^ { 2 } + \left( z _ { 2 } - z _ { 1 } \right) ^ { 2 } } ]$
and $PB^2 = (-1 - x)^2 + (3 - y)^2 + (-7 - z)^2$^
By the given condition, $PA^2 + PB^2 = 2k^2,$ we have
$(3 - x)^2 + (4 - y)^2 + (5 - z)^2 + (-1 - x)^2 + (3 - y)^2 + (-7 - z)^2 = 2k^2$
$\Rightarrow x^2 + 9 - 6x + y^2 +16 - 8y + z^2 + 25 - 10z+ x^2 + 2x + 1 + y^2 + 9 - 6y + z^2 + 49 + 14z = 2k^2$
$\Rightarrow 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = 2k^2 - 109$

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Question 53 Marks

Are the points $A(3, 6, 9), B(10, 20, 30)$ and $C(25, -41, 5),$ the vertices of a right-angled triangle$?$

Answer

Given: $A(3, 6, 9), B(10, 20, 30)$ and $C(25, -41, 5),$
According to the distance formula, we have
$AB^2 = (10 - 3)^2 + ( 20 - 6)^2 + (30 - 9)^2$
$= 49 + 196 + 441 = 686$
$BC^2 = (25 - 10)^2 + (-41 - 20)^2 + (5 - 30)^2$
$= 225 + 3721 + 625 = 4571$
and $CA^2 = (3 - 25)^2 + (6 + 41)^2 + (9 - 5)^2$
$= 484 + 2209 + 16 = 2709$
We observe that, $CA^2 + AB^2 \neq BC^2$
Hence, the $\triangle ABC$ is not a right angled triangle.

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Question 63 Marks

Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear.

Answer

We know that points are said to be collinear if they lie on a line
Now, $\mathrm{PQ}=\sqrt{(1+2)^{2}+(2-3)^{2}+(3-5)^{2}}=\sqrt{9+1+4}=\sqrt{14}$
$\mathrm{QR}=\sqrt{(7-1)^{2}+(0-2)^{2}+(-1-3)^{2}}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14}$
and $\mathrm{PR}=\sqrt{(7+2)^{2}+(0-3)^{2}+(-1-5)^{2}}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14}$
Thus, PQ + QR = PR
Hence, P, Q and R are collinear.

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Question 73 Marks

Show that the points A (1, 2, 3), B(-1, -2, -1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.

Answer

To show ABCD is a parallelogram we need to show opposite side are equal

Note that
$AB = \sqrt {{{(-1 - 1)}^2} + {{( - 2 - 2)}^2} + {{(-1 - 3)}^2}}$$ = \sqrt {4 + 16 + 16} = \sqrt {36} = 6$
$BC = \sqrt {{{(2 + 1)}^2} + {{( 3 + 2)}^2} + {{(2 + 1)}^2}}$$= \sqrt {9 + 25 + 9} = \sqrt {43}$
$CD = \sqrt {{{(4 - 2)}^2} + {{( 7 -3)}^2} + {{(6-2)}^2}}$$ = \sqrt {4 + 16 + 16} = \sqrt {36} = 6$
$DA = \sqrt {{{(1-4)}^2} + {{( 2-7)}^{^2}} + {{(3-6)}^2}}$$= \sqrt {9 + 25 + 9} = \sqrt 43$

Since AB = CD and BC = AD, ABCD is a parallelogram.

Now it is required to prove that ABCD is not a rectangle. For this, we show that diagonals AC and BD are unequal. We have
$AC = \sqrt {{{(2-1)}^2} + {{( 3 - 2)}^2} + {{(2-3)}^2}}$ $= \sqrt {1 + 1 + 1} = \sqrt 3$
$BD = \sqrt {{{(24+1)}^2} + {{( 7 + 2)}^2} + {{(6+1)}^2}}$ $ = \sqrt {25 + 81 + 49} = \sqrt {155}$
Since AB$\ne$BD, ABCD is not a rectangle

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Question 83 Marks

Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the YZ-plane.

Answer

Let YZ-plane divides the line segment joining the points A(4, 8, 10) and B(6, 10, -8) at P(x, y, z) in the ratio k: 1. Then, the coordinates of P are
$\left( \frac { 4 + 6 k } { k + 1 } , \frac { 8 + 10 k } { k + 1 } , \frac { 10 - 8 k } { k + 1 } \right)$
$\left[ \begin{array} { l } { \because \text { coordinates of internal division, } } \\ { \left( \frac { m _ { 1 } x _ { 2 } + m _ { 2 } x _ { 1 } } { m _ { 1 } + m _ { 2 } } , \frac { m _ { 1 } y _ { 2 } + m _ { 2 } y _ { 1 } } { m _ { 1 } + m _ { 2 } } , \frac { m _ { 1 } z _ { 2 } + m _ { 2 } z _ { 1 } } { m _ { 1 } + m _ { 2 } } \right) } \end{array} \right]$
Since P lies on the YZ-plane, its x-coordinate is zero,
i.e., $\frac { 4 + 6 k } { k + 1 } = 0 \quad \Rightarrow \quad k = - \frac { 2 } { 3 }$
Therefore, YZ-plane divides AB externally in the ratio 2:3.

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