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Maths NCERT Exemplar Solutions Class 12th Chapter Ten

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Three Dimensional Geometry

If two distinct points Q, R lie on the line of intersection of the planes x + 2y – z = 0 and 3x 5y + 2z = 0 and PQ = PR = $\sqrt[]{18}$ where the point P is (1, 2, 3), then the area of the triangle PQR is equal to

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Payal Gupta

Contributor-Level 10

Let vector along L is $\vec{x}$

$\vec{x} = | \begin{matrix} i & j & k \\ - 1 & 2 & - 1 \\ 3 & - 5 & 2 \end{matrix} |$

= $- \hat{i} - \hat{j} - \hat{k}$

Area of $Δ P Q R = \frac{1}{2} | \vec{P Q} * \vec{P R} |$

$= \frac{1}{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & - 1 \\ \frac{- 5}{3} & \frac{4}{3} & - \frac{11}{3} \end{matrix} |$

$= \frac{4}{3} \sqrt[]{38}$

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Straight Lines

Let the mirror image of a circle c₁ : x² + y² – 2x – 6y + a = 0 in line y = x + 1 be c₂ : 5x² + 5y² + 10gx + 10fy + 38 = 0. If r is the radius of circle c₂, then a + 6r² is equal to………….

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Vishal Baghel

Contributor-Level 10

${(x - 1)}^{2} + {(y - 3)}^{2} = 10 - α$

$\begin{array}{l} x - y = - 1 \\ x + y = a + b \end{array}} \to m (\frac{a + b - 1}{2}, \frac{a + b + 1}{2})$

A' = 2m – A = (b – 1, a + 1)

$C_{2} : x^{2} + y^{2} + 2 g x + 2 f y + \frac{38}{5} = 0$

r $\sqrt[]{g^{2} + f^{2} - c}$

$= \sqrt[]{4 + 4 - \frac{38}{5}} = \sqrt[]{\frac{2}{5}}$

$C_{1} : \sqrt[]{\frac{2}{5}} = \sqrt[]{1 + 9 - α} = \sqrt[]{10 - α}$

$α + 6 r^{2} = \frac{48}{5} + 6 (\frac{2}{5}) = \frac{60}{5} = 12$

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Vector Algebra

Let $\hat{a} a n d \hat{b}$ be two unit vectors such that the angle between them is $\frac{π}{4} .$ If is the angle between the vectors $(\hat{a} + \hat{b}) a n d (\hat{a} + 2 \hat{b} + 2 (\hat{a} * \hat{b})),$ then the value of 164cos2 is equal to:

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Vishal Baghel

Contributor-Level 10

$(\hat{a} + \hat{b}) . (\hat{a} + 2 \hat{b} + 2 (\hat{a} * \hat{b})), \hat{a} . \hat{b} = \frac{1}{\sqrt[]{2}}$

= 1 + $\frac{1}{\sqrt[]{2}} + \sqrt[]{2}$ + 2 + 0 + 0

$= 3 + \frac{3}{\sqrt[]{2}}$

$| \hat{a} + 2 \hat{b} + 2 {(\hat{a} * \hat{b})}^{2} | = 1 + 4 + 4 (\frac{1}{2}) + 2 \sqrt[]{2} + 0 = 7 + 2 \sqrt[]{2}$

$\Rightarrow (2 + \sqrt[]{2}) (7 + 2 \sqrt[]{2}) c o s^{2} θ = 9 + \frac{9}{2} + 9 \sqrt[]{2} = \frac{27 + 18 \sqrt[]{2}}{2}$

$\Rightarrow c o s^{2} θ = \frac{27 + 18 \sqrt[]{2}}{2} * \frac{1}{18 + 11 \sqrt[]{2}} = \frac{(27 + 18 \sqrt[]{2}) (18 - 11 \sqrt[]{2})}{2 (324 - 242)}$

$= \frac{486 + 324 \sqrt[]{2} - 297 \sqrt[]{2} - 396}{2 * 82}$

164 cos² (θ)= 90 + $2$ $7$

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Straight Lines

Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1, 1). If the line AP intersects the line BC at the point Q(k1, k2), then k1 + k2 is equal to:

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Vishal Baghel

Contributor-Level 10

${(a - 1)}^{2} + 4 = {(b - 1)}^{2} + 16$

$= {(a - 1)}^{2} + {(b - 1)}^{2}$

${(b - 1)}^{2} = 4 & {(a - 1)}^{2} = 16$

$\Rightarrow b = 1 \pm 2 a = 1 \pm 4$

= 3, 1= 5, 3

x + 2y = 3 …… (i)

3x – y = 8…… (ii)

x + 2y = 3

3x – y = 83 * 2

$\oplus \to 7 x = - 13 \Rightarrow x = \frac{- 13}{7}$

$y = \frac{- 39}{7} + 8 = \frac{17}{7}$

$k_{1} + k_{2} = x + y = \frac{4}{7}$

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Vector Algebra

Let $\vec{a} = 3 \hat{i} + \hat{j} a n d \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} .$ Let $\vec{c}$ be vector satisfying $\vec{a} * (\vec{b} * \vec{c}) = \vec{b} + λ \vec{c} .$ If $\vec{b} a n d \vec{c}$ are non-parallel, then the value of $λ$ is:

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Vishal Baghel

Contributor-Level 10

$(\overset{?}{a} . \overset{?}{c}) \overset{?}{b} ? (\overset{?}{a} . \overset{?}{b}) \overset{?}{c}$

$= \overset{?}{b} + ? \overset{?}{c} (g i v e n)$

$\overset{?}{a} . \overset{?}{c} = 1, ? = ? \overset{?}{a} . \overset{?}{b}$

$? = ? (3, 1, 0) . (1, 2, 1)$

= (3 + 2)= 5

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Straight Lines

Let a line L pass through the point of intersection of the lines bx + 10y – 8 = 0 and $2 x - 3 y = 0, b \in R - {\frac{4}{3}} .$ If the line L also passes through the point (1, 1) and touches the circle 17(x2 + y2) = 16, then the eccentricity of the ellipse $\frac{x^{2}}{5} + \frac{y^{2}}{b^{2}} = 1$ is:

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Vishal Baghel

Contributor-Level 10

y – 1 = m(x – 1)

mx – y + 1 – m = 0

$\frac{| 1 - m |}{\sqrt[]{m^{2} + 1}} = \frac{4}{\sqrt[]{17}}$

$17 {(1 - m)}^{2} = 16 (m^{2} + 1)$

m² – 3m + 1 = 0

$\begin{array}{l} b x + 10 y - 8 = 0 \\ y = \frac{2}{3} x \end{array}} \Rightarrow (b + \frac{20}{3}) x = 8$

$x = \frac{24}{3 b + 20}, y = \frac{16}{3 b + 20}$

b = $\frac{2 (3 - 2 \sqrt[]{2}) (4 + 3 \sqrt[]{2})}{- 2}$

$= \frac{12 - 8 \sqrt[]{2} + 9 \sqrt[]{2}}{- 1} = - \sqrt[]{2}$

b² = 2

$\frac{x^{2}}{5} + \frac{y^{2}}{2} = 1$

2 = 5 (1 – e²)

$e^{2} = \frac{3}{5}$

$e$ $=$

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Straight Lines

Let $P Q$ be a diameter of the circle $x^{2} + y^{2} = 9$ . If $α$ and $β$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x + y = 2$ respectively, then the maximum value of $α β$ is.

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alok kumar singh

Contributor-Level 10

$D = |\begin{matrix} 1 & - 2 & 3 \\ 2 & 1 & 1 \\ 1 & - 7 & a \end{matrix}| = 0 \Rightarrow a = 8$

also, $D_{1} = |\begin{matrix} 9 & - 2 & 3 \\ b & 1 & 1 \\ 24 & - 7 & 8 \end{matrix}| = 0 \Rightarrow b = 3$

hence, $a - b = 8 - 3 = 5$

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Vector Algebra

If $\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ , then the value of $| \hat{i} * (\vec{a} * \hat{i}) |^{2} + | \hat{j} * (\vec{a} * \hat{j}) |^{2} + | \hat{k} * (\vec{a} * \hat{k}) |^{2}$ is equal to

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alok kumar singh

Contributor-Level 10

Given ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{r} \frac{2^{10} {(2 x^{2} + 3 x + 4)}^{10}}{x^{20}} = \sum_{r = 0}^{20} ? ? \frac{a_{r} 2^{r}}{x^{r}} \\ \Rightarrow 2^{10} \sum_{r = 0}^{20} ? ? a_{r} x^{r} = \sum_{r = 0}^{20} ? ? a_{r} 2^{r} x^{(20 - r)} (from (i)) \end{array}$

now, comparing coefficient of $x^{7}$ from both sides

(take $r = 7$ in L.H.S. and $r = 13$ in R.H.S.)

$2^{10} a_{7} = a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}} = 2^{3} = 8$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Straight Lines

If the perpendicular bisector of the line segment joining the points $P (1,4)$ and $Q (k, 3)$ has $y$ -intercept equal to -4 , then a value of $k$ is:

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alok kumar singh

Contributor-Level 10

$m a x {n (A), n (B)} \leq n (A \cup B) \leq n (U)$

$\Rightarrow 76 \leq 76 + 63 - x \leq 100$

$\Rightarrow - 63 \leq - x \leq - 39$

$\Rightarrow 63 \geq x \geq 39$

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Mechanical Properties of Fluids

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be

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Payal Gupta

Contributor-Level 10

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

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