The distance of point P (3, 4, 5)
Question: The distance of point P (3, 4, 5) from the y z-plane is (A) 3 units (B) 4 units (C) 5 units (D) 550 Solution: (A) 3 units Explanation: From basic ides of three-dimensional geometry, we know that x-coordinate of a point is its distance from y z plane. Distance of Point P (3, 4, 5) from y z plane is given by its x coordinate. ∵x-coordinate of point P = 3 Distance of (3, 4, 5) from y z plane is 3 units Hence, option (A) is the only correct choice....
Read More →Find the value
Question: Find the (i) lengths of the axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity and (v) length of the rectum of each of the following the hyperbola : $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ Solution: Given Equation: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ Comparing with the equation of hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ we get, a = 3 and b = 4 (i) Length of Transverse axis = 2a = 6 units. Length of Conjugate axis = 2b = 8 units. (ii) Co...
Read More →What are the coordinates of the vertices of a
Question: What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin? Solution: Given thatcube with 2 units edge, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin. The Coordinates of the vertices are; (0,...
Read More →Prove that the points (0, – 1, – 7), (2, 1, – 9)
Question: Prove that the points (0, 1, 7), (2, 1, 9) and (6, 5, 13) are collinear. Find the ratio in which the first point divides the join of the other two. Solution: Given that three points $A(0,-1,-7), B(2,1,-9)$ and $C(6,5,-13)$ are collinear Therefore we can write as $\mathrm{AB}=\sqrt{(2-0)^{2}+(1-(-1))^{2}+((-9)-(-7))^{2}}=\sqrt{4+4+4}=2 \sqrt{3}$ $\mathrm{BC}=\sqrt{(6-2)^{2}+(5-1)^{2}+((-13)-(-9))^{2}}=\sqrt{16+16+16}=4 \sqrt{3}$ $\mathrm{AC}=\sqrt{(6-0)^{2}+(5-(-1))^{2}+((-13)-(-7))^{2}...
Read More →Manufacturer can sell x items at a price of rupees
Question: Manufacturer can sell $x$ items at a price of rupees $\left(5-\frac{x}{100}\right)$ each. The cost price is Rs $\left(\frac{x}{5}+500\right) .$ Find the number of items he should sell to earn maximum profit. Solution: Profit =S.P. - C.P. $\Rightarrow P=x\left(5-\frac{x}{100}\right)-\left(500+\frac{x}{5}\right)$ $\Rightarrow P=5 x-\frac{x^{2}}{100}-500-\frac{x}{5}$ $\Rightarrow \frac{d P}{d x}=5-\frac{x}{50}-\frac{1}{5}$ For maximum or minimum values of $P$, we must have $\frac{d P}{d x...
Read More →The mid-point of the sides of a triangle are (1, 5, – 1),
Question: The mid-point of the sides of a triangle are (1, 5, 1), (0, 4, 2) and (2, 3, 4). Find its vertices. Also find the centroid of the triangle. Solution: Given the mid-point of the sides of a triangle are $(1,5,-1),(0,4,-2)$ and $(2,3,$, 4). Let the vertices be $A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right)$ and $C\left(x_{3}, y_{3}, z_{3}\right)$ respectively. Now by using midpoint formula we get $\Rightarrow\left(\frac{\mathrm{X}_{1}+\mathrm{x}_{2}}{2}, \frac{\math...
Read More →The total cost of producing x radio sets per day
Question: The total cost of producing $x$ radio sets per day is Rs $\left(\frac{x^{2}}{4}+35 x+25\right)$ and the price per set at which they may be sold is Rs. ( $50-\frac{x}{2}$ ). Find the daily Solution: Profit =S.P. - C.P. $\Rightarrow P=x\left(50-\frac{x}{2}\right)-\left(\frac{x^{2}}{4}+35 x+25\right)$ $\Rightarrow P=50 x-\frac{x^{2}}{2}-\frac{x^{2}}{4}-35 x-25$ $\Rightarrow \frac{d P}{d x}=50-x-\frac{x}{2}-35$ For maximum or minimum values of $P$, we must have $\frac{d P}{d x}=0$ $\Righta...
Read More →Find the maximum slope of the curve
Question: Find the maximum slope of the curve $y=-x^{3}+3 x^{2}+2 x-27 .$ Solution: Given : $y=-x^{3}+3 x^{2}+2 x-27$ .....(1) Slope $=\frac{d y}{d x}=-3 x^{2}+6 x+2$ Now, $M=-3 x^{2}+6 x+2$ $\Rightarrow \frac{d M}{d x}=-6 x+6$ For maximum or minimum values of $M$, we must have $\frac{d M}{d x}=0$ $\Rightarrow-6 x+6=0$ $\Rightarrow 6 x=6$ $\Rightarrow x=1$ Substituing the value of $x$ in eq. (1), we get $y=-1^{3}+3 \times 1^{2}+2 \times 1-27=-23$ $\frac{d^{2} M}{d x^{2}}=-60$ So, the slope is ma...
Read More →Show that the three points A (2, 3, 4),
Question: 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. Solution: Given three points are $A(2,3,4), B(-1,2,-3)$ and $C(-4,1,-10)$ To find collinear points, $\mathrm{AB}=\sqrt{(2+1)^{2}+(3-2)^{2}+(4+3)^{2}}$ $=\sqrt{9+1+49}$ $=\sqrt{59}$ Now consider, $B C=\sqrt{(-1+4)^{2}+(2-1)^{2}+(-3+10)^{2}}$ $=\sqrt{9+1+49}$ $=\sqrt{59}$ Again we have, $\mathrm{AC}=\sqrt{(2+4)^{2}+(3-1)^{2}+(4+10)^{2}}$ $=\sqrt{36+4+196}$ $=\sqrt...
Read More →Find the eccentricity of an ellipse whose latus rectum is one half of its major axis.
Question: Find the eccentricity of an ellipse whose latus rectum is one half of its major axis. Solution: Let the equation of the required ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$(i) It is given that, Length of Latus Rectum $=\frac{1}{2}$ major Axis We know that, Length of Latus Rectum $=\frac{2 b^{2}}{a}$ and Length of Minor Axis $=2 \mathrm{a}$ So, according to the given condition, $\frac{2 b^{2}}{a}=\frac{1}{2} \times 2 a$ $\Rightarrow \frac{2 b^{2}}{a}=a$ $\Rightarrow 2 b^{2}=a...
Read More →Let A (2, 2, – 3), B (5, 6, 9) and C (2, 7, 9) be the vertices of a triangle.
Question: Let A (2, 2, 3), B (5, 6, 9) and C (2, 7, 9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point D. Find the coordinates of D. Solution: Given $A(2,2,-3), B(5,6,9)$ and $C(2,7,9)$ are the vertices of a triangle. And also given that the internal bisector of the angle $A$ meets $B C$ at the point D. $A B=\sqrt{(5-2)^{2}+(6-2)^{2}+(9-(-3))^{2}}=\sqrt{9+16+144}=\sqrt{169}=13$ Now, $\mathrm{AC}=\sqrt{(2-2)^{2}+(7-2)^{2}+(9-(-3))^{2}}=\sqrt{0+25+144}=\sq...
Read More →Find the eccentricity of an ellipse whose latus rectum is one half of its
Question: Find the eccentricity of an ellipse whose latus rectum is one half of its minor axis. Solution: Let the equation of the required ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ ............(i) It is given that Length of Latus Rectum $=\frac{1}{2}$ minor Axis We know that, Length of Latus Rectum $=\frac{2 b^{2}}{a}$ and Length of Minor Axis = 2b So, according to the given condition, $\frac{2 b^{2}}{a}=\frac{1}{2} \times 2 b$ $\Rightarrow \frac{2 b^{2}}{a}=b$ $\Rightarrow \frac{2 ...
Read More →Find the point on the curvey
Question: Find the point on the curvey $y^{2}=2 x$ which is at a minimum distance from the point $(1,4)$. Solution: Suppose a point $(\mathrm{x}, \mathrm{y})$ on the curve $y^{2}=2 x$ is nearest to the point $(1,4) .$ Then, $y^{2}=2 x$ $\Rightarrow x=\frac{y^{2}}{2}$ ....(1) $d^{2}=(x-1)^{2}+(y-4)^{2}$ [Using distance formula] Now, $Z=d^{2}=(x-1)^{2}+(y-4)^{2}$ $\Rightarrow Z=\left(\frac{y^{2}}{2}-1\right)^{2}+(y-4)^{2}$ [From eq. (1)] $\Rightarrow Z=\frac{y^{4}}{4}+1-y^{2}+y^{2}+16-8 y$ $\Right...
Read More →If the origin is the centroid of a triangle
Question: If the origin is the centroid of a triangle ABC having vertices A (a, 1, 3), B ( 2, b, 5) and C (4, 7, c), find the values of a, b, c. Solution: Given triangle $A B C$ having vertices $A(a, 1,3), B(-2, b,-5)$ and $C(4,7, c)$ and origin is the centroid. For a triangle the coordinates of the centroid is given by the average of the coordinates of its vertices. Therefore, $\Rightarrow(0,0,0)=\left(\frac{a+(-2)+4}{3}, \frac{1+b+7}{3}, \frac{3+(-5)+c}{3}\right)$ Now by comparing the each poi...
Read More →Find the coordinate of the points which trisect
Question: Find the coordinate of the points which trisect the line segment joining the points A (2, 1, 3) and B (5, 8, 3). Solution: Given the line segment joining the points are $A(2,1,-3)$ and $B(5,-8,3)$. Now let $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\mathrm{Q}\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ be the points which trisects the line segment. $\Rightarrow \mathrm{P}$ divides $\mathrm{AB}$ in the ratio $2: 1$ $\Rightarrow x_{1}=\fr...
Read More →Find the coordinates of a point on the parabola
Question: Find the coordinates of a point on the parabola $y=x^{2}+7 x+2$ which is closest to the strainght line $y=3 x-3$. Solution: Let coordinates of the point on the parabola be $(x, y)$. Then, $y=x^{2}+7 x+2$ ......(1) Let the distance of a point $\left(x,\left(x^{2}+7 x+2\right)\right)$ from the line $y=3 x-3$ be $S .$ Then, $S=\left|\frac{-3 x+\left(x^{2}+7 x+2\right)+3}{\sqrt{10}}\right|$ $\Rightarrow \frac{d S}{d t}=\frac{-3+2 x+7}{\sqrt{10}}$ For maximum or minimum values of $S$, we mu...
Read More →Find the equation of an ellipse whose eccentricity is
Question: Find the equation of an ellipse whose eccentricity is $\frac{2}{3}$, the latus rectum is 5, and the center is at the origin. Solution: Let the equation of the required ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (i) Given that Eccentricity $=\frac{2}{3}$ we know that, Eccentricity, $e=\frac{c}{a}$ $\Rightarrow \frac{2}{3}=\frac{c}{a}$ $\Rightarrow c=\frac{2}{3} a$ We know that, $c^{2}=a^{2}-b^{2}$ $\Rightarrow\left(\frac{2 a}{3}\right)^{2}=a^{2}-b^{2}$ $\Rightarrow \frac{4 a...
Read More →Three vertices of a Parallelogram ABCD
Question: Three vertices of a Parallelogram ABCD are A (1, 2, 3), B ( 1, 2, 1) and C (2, 3, 2). Find the fourth vertex D. Solution: Given three consecutive vertices of a parallelogram $A B C D$ are $A(1,2,3), B(-1$, $-2,-1)$ and $C(2,3,2)$ Let the fourth vertex be $D(x, y, z)$. Now by using midpoint formula, Midpoint of diagonal $A C=P\left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right)=P\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$ Midpoint of diagonal BD $=\mathrm{P}\left(\frac{\math...
Read More →Find the point on the parabolas
Question: Find the point on the parabolas $x^{2}=2 y$ which is closest to the point $(0,5)$. Solution: Let the required point be $(x, y)$. Then, $x^{2}=2 y$ $\Rightarrow y=\frac{x^{2}}{2}$ ....(1) The distance between points $(x, y)$ and $(0,5)$ is given by $d^{2}=(x)^{2}+(y-5)^{2}$ Now, $d^{2}=Z$ $\Rightarrow Z=(x)^{2}+\left(\frac{x^{2}}{2}-5\right)^{2}$ $\Rightarrow Z=x^{2}+\frac{x^{4}}{4}+25-5 x^{2}$ $\Rightarrow \frac{d Z}{d y}=2 x+x^{3}-10 x$ For maximum or a minimum values of $Z$, we must ...
Read More →The mid-points of the sides of a triangle are
Question: The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, 1). Find its vertices. Solution: Given the mid-points of the sides of a triangle are $(5,7,11),(0,8,5)$ and $(2,3,-$ 1). Let he vertices be $A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right)$ and $A\left(x_{3}, y_{3}, z_{3}\right)$ respectively. Using midpoint formula, $\Rightarrow\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}, \frac{\mathrm{z}_{1}+\...
Read More →Find the point on the curve
Question: Find the point on the curve $x^{2}=8 y$ which is nearest to the point $(2,4)$. Solution: Let $(x, y)$ be nearest to the point $(2,4)$. Then, $x^{2}=8 y$ $\Rightarrow y=\frac{x^{2}}{8}$ .....(1) $d^{2}=(x-2)^{2}+(y-4)^{2}$ [Using distance formula] Now, $Z=d^{2}=(x-2)^{2}+(y-4)^{2}$ $\Rightarrow Z=(x-2)^{2}+\left(\frac{x^{2}}{8}-4\right)^{2}$ [From eq. (1)] $\Rightarrow Z=x^{2}+4-4 x+\frac{x^{4}}{64}+16-x^{2}$ $\Rightarrow \frac{d Z}{d y}=-4+\frac{4 x^{3}}{64}$ For maximum or minimum val...
Read More →Find the centroid of a triangle,
Question: Find the centroid of a triangle, the mid-point of whose sides are D (1, 2, 3), E (3, 0, 1) and F ( 1, 1, 4). Solution: Given: Mid-points of sides of triangle DEF are: $\mathrm{D}(1,2,-3), \mathrm{E}(3,0,1)$ and $\mathrm{F}(-1,1,-4)$ By using the geometry of centroid, We know that the centroid of triangle DEF is given as: $\mathrm{G}=[(1+3-1) / 3,(2+0+1) / 3,(-3+1-4) / 3]$ $=(1,1,-2)$ Hence, the centroid of triangle DEF is $(1,1,-2)$....
Read More →Find the third vertex of triangle
Question: Find the third vertex of triangle whose centroid is origin and two vertices are (2, 4, 6) and (0, 2, 5). Solution: Given the centroid is origin and two vertices are $(2,4,6)$ and $(0,-2,-5)$. Let the third vertex be $(x, y, z)$ For a triangle the coordinates of the centroid is given by the average of the coordinates of its vertices. $\Rightarrow(0,0,0)=\left(\frac{2+0+x}{3}, \frac{4+(-2)+y}{3}, \frac{6+(-5)+z}{3}\right)$ $\Rightarrow \frac{2+\mathrm{x}}{3}=0, \therefore \mathrm{x}=-2$ ...
Read More →Find the point on the curve
Question: Find the point on the curve $y 2=4 x$ which is nearest to the point $(2,-8)$ Solution: Let point $(x, y)$ be the nearest to the point $(2,-8)$. Then, $y^{2}=4 x$ $\Rightarrow x=\frac{y^{2}}{4}$ ......(1) $d^{2}=(x-2)^{2}+(y+8)^{2}$ [Using distance formula] Now, $Z=d^{2}=(x-2)^{2}+(y+8)^{2}$ $\Rightarrow Z=\left(\frac{y^{2}}{4}-2\right)^{2}+(y+8)^{2}$ [From eq. (1)] $\Rightarrow Z=\frac{y^{4}}{16}+4-y^{2}+y^{2}+64+16 y$ $\Rightarrow \frac{d Z}{d y}=\frac{4 y^{3}}{16}+16$ For maximum or ...
Read More →Show that the triangle ABC with vertices A (0, 4, 1),
Question: Show that the triangle ABC with vertices A (0, 4, 1), B (2, 3, 1) and C (4, 5, 0) is right angled. Solution: Given vertices are $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$. To prove right angled triangle, consider $\mathrm{AB}=\sqrt{(0-2)^{2}+(4-3)^{2}+(1+1)^{2}}=\sqrt{4+1+4}=\sqrt{9}=3$ $B C=\sqrt{(2-4)^{2}+(3-5)^{2}+(-1-0)^{2}}=\sqrt{4+4+1}=\sqrt{9}=3$ $\mathrm{AC}=\sqrt{(0-4)^{2}+(4-5)^{2}+(1-0)^{2}}=\sqrt{16+1+1}=\sqrt{18}=3 \sqrt{2}$ $\Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{B...
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