Solve the Following Question.(3 Marks) - Maths STD 11 Questions - Vidyadip

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Solve the Following Question.(3 Marks)

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Find the ratio in which the line Segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z =5.

Answer

(2, -1, 3) and (-1, 2, 1)
x + y + z = 5
Assume plane divides line in ratio $\lambda : 1$
so point P which is diving line in $\lambda : 1$ ratio is
$\text{P}=\Big(\frac{-\lambda+2}{\lambda+1},\frac{2\lambda-1}{\lambda+1},\frac{\lambda+3}{\lambda+1}\Big)$
P lies on plane x + y + z = 5
$-\lambda+2+2\lambda-1+\lambda+3=5\lambda+5$
$3\lambda=-1\Rightarrow\lambda=-1:3$
So plane diving line in 1 : 3 ratio externally

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Question 23 Marks

Planes are drawn parallel to the coordinate planes through the points (3, 0, -1) and (-2, 5, 4).
Find the lengths of the edges of the parallelopiped so formed.

Answer

IMAGE

Let P ≡ (3, 0, -1), Q ≡ (-2, 5, 4)

PE = Distance between the parallel planes ABCP and FQDE

= |4 + 1| = 5 (These planes are perpendicular to the z-axis)

PA = Distance between the parallel planes ABQF and PCDE

= |-2 - 3| = 5 (These planes are perpendicular to the x-axis)

Similarly, PC = |5 - 0| = 5

Thus, the length of the edges of the parallelepiped are 5, 5 and 5

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Question 33 Marks

Find the ratio in which the sphere $x^2 + y^2 + z^2 = 504$ divides the line joining the points $(12, -4, 8)$ and $(21, -9, 18).$

Answer

$(12, -4, 8)$ and $(27, -9, 18)$
Assume point $P$ is dividing line in $\lambda: 1$ ratio, we get
$\text{P}=\Big(\frac{27\lambda+12}{\lambda+1},\frac{-9\lambda+-4}{\lambda+1},\frac{18\lambda+8}{\lambda+1}\Big)$
$P$ lies on Sphere, so substitute in Sphere equation
$x^2 + y^2 + z^2 = 504$
$9(9\lambda+4)^2+(9\lambda+4)^2 +4(9\lambda+4)^2=504(\lambda+1)^2$
$729\lambda^2 +81\lambda^2 +324\lambda^2 +648\lambda+72\lambda +288\lambda+144 +16+64 = 504\lambda^2 +1008\lambda +504$
$(1134-504) \lambda^2 +(1008-1008)\lambda+224 – 504 = 0$
$630\lambda^2 = 280$
$\lambda^2=\frac{4}{9}$
$\lambda=2:3$

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Question 43 Marks

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle $\angle\text{BAC}$ meets BC.

Answer

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3)

Angle bisctor at A divides BC in ratio of AB : AC

AB $=\sqrt{1+4+4}=3$

AC $=\sqrt{14+9+36}=7$

Assume D divides BC

m : n = 3 : 7

so D $=\Big(\frac{-3}{10},\frac{25}{10},\frac{-2}{10}\Big)$

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Question 53 Marks

The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, -5, 7) and (-1, 7, -6) respectively, find the coordinates of the point C.

Answer

Given Centroid (1, 1, 1)

A(3, -5, 7) and B(-1, 7, -6)

Equating terms, we get

$1=\frac{3-1+\text{x}_3}{3}$

$1=\frac{-5+7+\text{y}_3}{3}$

$1=\frac{7-6+\text{z}_3}{3}$

$(\text{x}_3,\text{y}_3,\text{z}_3)=(1, 1, 2)$

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Question 63 Marks

The mid-points of the sides of a triangle $ABC$ are given by $(-2, 3, 5), (4, -1, 7)$ and $(6, 5, 3).$ Find the coordinates of $A, B$ and $C.$

Answer

Given mid-points $D(-2, 3, 5), E(4, -1, 7)$ and $F(6, 5, 3)$
Assume $D$ is mid-point of $AB, E$ is mid-point of $BC$
$F$ is mid-point of $CA$
$A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$
From mid-point formula, we get following equations
$x_1+ x_2 = -4; x_2 + x_3= 8; x_3 + x_1 = 12$
$y_1+ y_2 = 6; y_2 + y_3= -2; y_3 + y_1 = 10$
$z_1+ z_2 = 10; z_2 + z_3= 14; z_3 + z_1 = 6$
Solving above set of equations we get
$A = (0, 9, 1)$
$B = (4, -3, 9)$
$C = (12, 1, 5)$

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Question 73 Marks

Find the points on z-axis which are at a distance $\sqrt{21}$ from the point (1, 2, 3).

Answer

Let P(0, 0, z) be at a distance of $\sqrt{21}$ from Q(1, 2, 3)

So

$\text{PQ}=\sqrt{(0-1)^2+(0-2)^2+(\text{z}-3)^2}$

$\sqrt{21}=\sqrt{(1)^2+(2)^2+(\text{z}-3)^2}$

$21-5=(\text{z}-3)^2$

$16=(\text{z}-3)^2$

$\text{z}-3=\pm4$

$\text{z}-7\text{ and }\text{z}--1$

So, the required points are (0, 0, 7) and (0, 0, -1)

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Question 83 Marks

Show that the three points A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which C divides AB.

Answer

If points are collinear then all points lie on same line
and DR's should be proportional
A(2, 3, 4), B(-1, 2, -3) and C(4, 1, -10)
DR's of AB = (3, 1, 7)
DR's of BC = (3, 1, 7)
So A, B, C are collinear
Length of AC $=\sqrt{36+4+196}=\sqrt{236}$
Length of AB $=\sqrt{9+1+49}=\sqrt{59}$
Ratio is AC : AB = 2 : 1
So C divides AB in ratio 2 : 1 externally

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Question 93 Marks

Find the centroid of a triangle, mid-points of whose sides are $(1, 2, -3), (3, 0, 1)$ and $(-1, 1, -4).$

Answer

$(1, 2, -3), (3, 0, 1)$ and $(-1, 1, 4)$
Centroid of Traingle is given by
$\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3},\frac{\text{z}_1+\text{z}_2+\text{z}_3}{3}\Big)$
We know that
$x_1 + x_2 = 2$
$x_2 + x_3 = 6$
$x_1+ x_3 = -2$
Adding all gives $\Rightarrow 2(x_1 + x_2 + x_3) = 6$
so $x_1 + x_2 + x_3 = 3$
similarly, $y_1 + y_2 + y_3 = 3; z_1 + z_2 + z_3 = -6$
Centroid $= (1, 1, -2)$

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Question 103 Marks

Find the point on $y-$axis which is equidistant from the points $(3, 1, 2)$ and $(5, 5, 2).$

Answer

Let $P(0, y, 0)$ be a point on y-axis which is equidistant from $Q(3, 1, 2)$ and $R(5, 5, 2).$
So
$(PR)^2 = (PQ)^2 \Rightarrow (0 - 5)^2 + (y - 5)^2 + (0 - 2)^2 = (0 - 3) + (y - 1)^2 + (0 + 2)^2$
$\Rightarrow 25 + y^2 + 25 - 10y + 4 = 9 + y^2 + 1 - 2y + 4$
$\Rightarrow -10y + 2y = 14 - 54$
$\Rightarrow -14z - 8z = 16 - 49$
$\Rightarrow -8y = -40$
$\Rightarrow y = 5$
so, the required point is $(0, 5, 0)$

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Question 113 Marks

Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.

Answer

Let the triangle formed be $\triangle\text{ABC}$

$\text{(AB)}=\sqrt{(1-2)^2+(2-3)^2+(3-1)^2}$

$=\sqrt{(-1)^2+(-1)^2+(2)^2}$

$=\sqrt{6}\text{ units}$

$\text{BC}=\sqrt{(2-3)^2+(3-1)^2+(1-2)^2}$

$=\sqrt{(-1)^2+(2)^2+(-1)^2}$

$=\sqrt{6}\text{ units}$

$\text{AC}=\sqrt{(1-3)^2+(2-1)^2+(3-2)^2}$

$=\sqrt{(-2)^2+(1)^2+(1)^2}$

$=\sqrt{6}\text{ units}$

since, AB = BC = CA

So, $\triangle\text{ABC}$ is an equilateral $\triangle$

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Question 123 Marks

A point C with z-coordinate 8 lies on the line segment joining the points A(2, -3, 4) and B(8, 0, 10). Find its coordinates.

Answer

z-coordinate 8

A(2, -3, 4) and B(8, 0, 10)

DR's of AB = (6, 3, 6)

DR's of BC = (x - 8, y - 0, 8 - 10)

Given A, B, C lie on same line

So values of DR's should be proportional

$\frac{\text{x}-8}{6}=\frac{\text{y}}{3}=\frac{8-10}{6}$

So x = 6, y = -1

point is (6, -1, 8)

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Question 133 Marks

Find the distances of the point P(-4, 3, 5) from the coordinate axes.

Answer

IMAGE

Let PQ be the perpendicular to the xy-plane and QA be perpendicular from Q to the y-axis.

PA will be perpendicular to the x-axis

Also, QA = |3| and PQ = |5|

Now, distance of P from the x-axis:

$\text{PB}=\sqrt{\text{BQ}^2+\text{QP}^2}$

$=\sqrt{3^2+5^2}$

$=\sqrt{9+25}=\sqrt{34}$

Similarly,

From the right-angled $\triangle\text{PAQ.}$

distance of P from the y-axis:

$\text{PA}=\sqrt{\text{AQ}^2+\text{QP}^2}$

$=\sqrt{(-4)^2+(5)^2}$

$=\sqrt{16+25}=\sqrt{41}$

Similarly, the length of the perpendicular from P to the z-axis $=\sqrt{(-4)^2+(3)^2}$

$=\sqrt{16+9}$

$=\sqrt{25}=5$

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Question 143 Marks

If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divides AB.

Answer

A(3, 2, 4), B(9, 8, -10) and C(5, 4, -6)
$\text{AC}=\sqrt{4+4+4}=2\sqrt{3}$
$\text{AB}=\sqrt{36+36+36}=6\sqrt{3}$
$\text{BC}=\sqrt{16+16+16}=4\sqrt{3}$
$\text{AC}:\text{BC}=1:2$

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Question 153 Marks

A cube of side 5 has one vertex at the point(1, 0, -1) and the three edge from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.

Answer

Let P ≡ (1, 0, -1)
The length of each side of the cube is 5.
The three edges from vertex of the cube are drawn from P towards the
negative x and y axes and the positive z-axis.
Therefore, the coordinates of the vertex of the cube will be as follows:
x-coordinate = 1, 1-5 = -4, i.e. 1, -4

y-coordinate = 0, 0-5 = -5, i.e. 0, -5

z-coordinate = -1, -1 + 5 = 4, i.e. -1, 4

Hence, the remaining seven vertices of the cube are as follows:

(1, 0, 4)

(1, -5, -1)

(1, -5, 4)

(-4, 0, -1)

(-4, -5, 4)

(-4, -5, -1)

(4, 0, 4)

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Question 163 Marks

Determine the point on $z-$axis which is equidistant from the points $(1, 5, 7)$ and $(5, 1, -4).$

Answer

Let $P(0, 0, z)$ be the equidistant from $Q(1, 5, 7)$ and $R(5, 1, -4).$
So
$(PQ)^2 = (PR)^2 \Rightarrow (0 - 1)^2 + (0 - 5)^2 + (z - 7)^2 = (0 - 5) + (0 - 1)^2 + (z + 4)^2$
$\Rightarrow 1 + 25 + (z - 7)^2 = 25 + 1 + (z + 4)^2$
$\Rightarrow 26 + z^2 + 49 - 14z = 26 + z^2 + 8z + 16$
$\Rightarrow -14z - 8z = 16 - 49$
$\Rightarrow -22z = -33$
$\Rightarrow\text{z}=\frac{-33}{-22}$
$\Rightarrow\text{z}=\frac{3}{2}$
Required point $=\Big(0,\ 0,\ \frac{3}{2}\Big)$

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Question 173 Marks

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Answer

yz plane means x = 0

Given (2, 4, 5) and (3, 5, 4)

assume ratio to be m : n

lets equate x - term

$0=\frac{\text{3m+2n}}{\text{m+n}}$

3m = -2n

m : n = -2 : 3

which means yz plane divides the line in 2 : 3 ratio externally

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Question 183 Marks

The vertices of the triangle are $A(5, 4, 6), B(1, -1, 3)$ and $C(4, 3, 2)$.The intenal bisector of angle $A$ meets $BC$ at $D.$ Find the coordinates of $D$ and the length $AD.$

Answer

We know that angle bisector divides opposite side in ratio of other two sides
$\Rightarrow D$ divides $BC$ in ratio of $AB : AC$
$A(5, 4, 6), B(1, -1, 3)$ and $C(4, 3, 2)$
$\text{AB}\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2}$
$\text{AC}\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}$
$AB : AC = 5 : 3 = m : n$
$\text{D(x, y, z)}=\Big(\frac{\text{mx}_2+\text{nx}_1}{\text{m+n}},\frac{\text{my}_2+\text{ny}_1}{\text{m+n}},\frac{\text{mz}_2+\text{nz}_1}{\text{m+n}}\Big)$
Substitute values for $m : n = 5 : 3,$
$(x_1, y_1, z_1) = (1, -1, 3)$
$(x_2, y_2, z_2) = (4, 3, 2)$
$\text{D}=\Big(\frac{23}{8},\frac{3}{2},\frac{19}{8}\Big)$
$\therefore\ \text{AD}=\sqrt{\Big(5-\frac{23}{8}\Big)^2+\Big(4-\frac{12}{8}\Big)^2+\Big(6-\frac{19}{8}\Big)^2}$
$=\sqrt{\frac{17^2+20^2+29^2}{8^2}}$
$=\sqrt{\frac{289+400+841}{8^2}}$
$=\frac{\sqrt{1530}}{8}$

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