Let a line having direction ratios 1, -4, 2 intersect the lines x − 7 3 = y − 1 − 1 = z + 2 1 and x 2 = y − 7 3 = z 1 at the points A and B. Then (AB)2 is equal to _________.

x − 7 3 = y − 1 − 1 = z + 2 1 = r 1 A ( 3 r 1 + 7 , 1 − r 1 , − 2 + r 1 ) and x 2 = y − 7 3 = z 1 = r 2 B ( 2 r 2 , 7 + 3 r 2 , r 2 ) A / q , 3 r 1 − 2 r 2 + 7 1 = 3 r 2 + r 1 + 6 4 = r 1 − r 2 − 2 2 ⇒ r 1 = − 5 , r 2 = − 3 ∴ A ( − 8 , 6 , − 7 ) a n d B ( − 6 , − 2 , − 3 ) AB2 = 84 f ( x ) = { | 2 x 2 − 3 x − 7 | , x ≤ − 1 [ 4 x 2 − 1 ] , − 1 < x < 1 | x + 1 | + | x − 2 | , x ≥ 1 f(-1) = 1 f(1) = 3 Hence f(x) will be discontinuous at x = 1 and also 4x2 – 1 = 0 , 1 , 2 ⇒ x = ± 1 2 , ± 1 2 , ± 3 2

Let a line having direction ratios 1, -4, 2 intersect the lines $\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1}$ and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}$ at the points A and B. Then (AB)² is equal to _________.

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Alok Kumar Singh | Contributor-Level 10

8 months ago

$\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1} = r_{1}$

$A (3 r_{1} + 7, 1 - r_{1}, - 2 + r_{1})$

and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1} = r_{2}$

$B (2 r_{2}, 7 + 3 r_{2}, r_{2})$

$A / q, \frac{3 r_{1} - 2 r_{2} + 7}{1} = \frac{3 r_{2} + r_{1} + 6}{4} = \frac{r_{1} - r_{2} - 2}{2}$

$\Rightarrow r_{1} = - 5, r_{2} = - 3$

$∴ A (- 8, 6, - 7) a n d B (- 6, - 2, - 3)$

AB² = 84

$f (x) = {\begin{cases} | 2 x^{2} - 3 x - 7 |, x \leq - 1 \\ [4 x^{2} - 1], - 1 < x < 1 \\ | x + 1 | + | x - 2 |, x \geq 1 \end{cases}$

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x = \pm \frac{1}{2}, \pm \frac{1}{\sqrt[]{2}}, \pm \frac{\sqrt[]{3}}{2}$

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If $\int_{}^{} (\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$ 

Then the value of 'α' is equal to

Alok Kumar Singh

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{x^{2} c o s x}{1 + π^{2}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} {(\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) + (\frac{x^{2} c o s x}{1 + π^{- x}} + \frac{1 + s i n^{2} x}{1 + e^{- {(s i n x)}^{2023}}})} d x$

$= \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} (x^{2} c o s x + 1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} x^{2} c o s x d x + \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$ ....(1)

Let $I_{1} = \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x$

$I_{1} = \int_{0}^{\frac{π}{2}} 1 \cdot d x + \int_{0}^{\frac{π}{2}} (\frac{1 - c o s 2 x}{2}) d x$

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

$I_{1} = \frac{3 π}{4}$

Let $I_{2} = \int_{0}^{\frac{π}{2}} x^{2} c o s x d x$

$I_{2} = {[x^{2} (s i n x) - \int 2 x \int c o s x d x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} (s i n x) - 2 \int x s i n x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (x (- c o s x) + \int c o s x)]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (- x c o s x + s i n x)]}_{0}^{\frac{π}{2}}$

$I_{2} = (\frac{π^{2}}{4} - 2)$

Put l₁ and l₂ in (1)

$\frac{π^{2}}{4} - 2 + \frac{3 π}{4}$

$\frac{π^{2}}{4} + \frac{3 π}{4} - 2$

$\frac{π}{4} (π + 3) - 2$

α = 3

If $| \vec{a} | = 1$ , $| \vec{b} | = 4$ $\vec{a} \cdot \vec{b} = 2$  and $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$ Then the angle between $\vec{b}$ and $\vec{c}$ is

Alok Kumar Singh

Given $| \vec{a} | = 1$ , $| \vec{b} | = 4$ , $\vec{a} \cdot \vec{b} = 2$

$\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$

Dot product with $\vec{a}$ on both sides

$\vec{c} \cdot \vec{a} = - 6$ ... (1)

Dot product with $\vec{b}$ on both sides

$\vec{b} \cdot \vec{c} = - 48$ ... (2)

$\vec{c} \cdot \vec{c} = 4 {| \vec{a} \times \vec{b} |}^{2} + 9 {| \vec{b} |}^{2}$

${| \vec{c} |}^{2} = 4 [{| \vec{a} |}^{2} \cdot {| \vec{b} |}^{2} - {(\vec{a} \cdot \vec{b})}^{2}] + 9 {| \vec{b} |}^{2}$

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${| \vec{c} |}^{2} = 4 [12] + 144$

${| \vec{c} |}^{2} = 48 + 144$

${| \vec{c} |}^{2} = 192$

$c o s θ = \frac{\vec{b} \cdot \vec{c}}{| \vec{b} | | \vec{c} |}$

$c o s θ = \frac{- 48}{\sqrt[]{192} \cdot 4}$

$c o s θ = \frac{- 48}{8 \sqrt[]{3} \cdot 4}$

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Alok Kumar Singh

(a – 1) × 2 + (b – 2) × 5 + (g – 3) × 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

$λ = \frac{1}{10}$

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

$= 8 λ + 5 = \frac{8}{10} + 5 = 5.8$

Alok Kumar Singh

Take $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

$\vec{A B}$ = (α − 2) $\hat{i}$ + (β − 3) $\hat{j}$ + (γ − 4) $\hat{k}$

Now,

(α − 2) ⋅ 2 + (β − 3) ⋅3 + (γ − 4) ⋅ 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

⇒ 2α + 3β + 4γ = 29

Vishal Baghel

$L_{1} = \frac{x - λ}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z}{- \frac{1}{2}}$

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$∴ | λ | = 1$

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