Angle between line joining two vertices of a regular tetrahedron to centroid is: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

π - cos⁻¹(1/3)

π - cos⁻¹(2/3)

Maths NCERT Exemplar Solutions Class 11th Chapter Two Three Dimensional Geometry Ncert Solutions Maths class 12th Maths Class 12th

Angle between line joining two vertices of a regular tetrahedron to centroid is:

Option 1 -

π - cos⁻¹(2/3)

Option 2 -

π - cos⁻¹(1/3)

Option 3 -

2π/3

Option 4 -

3π/4

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 2

Detailed Solution:

$\overline{a} + \overline{b} + \overline{c} + \overline{d} = 0$

Magnitude of all vectors will be same as well as angle between these vectors will also be same. $\overline{a} + \overline{b} + \overline{c} = - \overline{d}$

$\begin{matrix} | \overline{a} |^{2} + | \overline{b} |^{2} + | \overline{c} |^{2} + 2 \overline{a} \cdot \overline{b} + 2 \overline{b} \cdot \overline{c} + 2 \overline{c} \cdot \overline{a} = | d |^{2} \\ 1 + 1 + 1 + 2 c o s ? α + 2 c o s ? α + 2 c o s ? α = 1 \\ 6 c o s ? α = - 2 \\ c o s ? α = - \frac{1}{3} \\ α = {c o s}^{- 1} ? (\frac{- 1}{3}) = π - {c o s}^{- 1} ? \frac{1}{3} \end{matrix}$

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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

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

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

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

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

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

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

$I_{1} = \frac{3 π}{4}$

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

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

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

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

$I_{2} = (\frac{π^{2}}{4} - 2)$

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

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

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

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

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

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If the foot of perpendicular from (1, 2, 3) to the line $\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z - 4}{1}$ is (a, b, g) then find a + b + g

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(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

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

$= 8 λ + 5 = \frac{8}{10} + 5 = 5.8$

If (α, β, γ) is mirror image of the point (2, 3, 4) with respect to the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ Then 2α + 3β + 4γ is

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

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

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

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