Maths NCERT Exemplar Solutions Class 12th Chapter Ten Exemplar Solution Maths Class 12th

If $\vec{a}, \vec{b}, \vec{c}$ **determine the vertices of a triangle, show that $\frac{1}{2} (\vec{b} * \vec{c} + \vec{c} * \vec{a} + \vec{a} * \vec{b})$ gives the $\vec{a}, \vec{b}, \vec{c}$ vector area of the triangle. Hence deduce the condition that the three points are collinear. Also, find the unit vector normal to the plane of the triangle.**

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Alok Kumar Singh | Contributor-Level 10

8 months ago

This is a Long Answer type Questions as classified in NCERT Exemplar

$\begin{array}{l} Since, \vec{a}, \vec{b} a n d \vec{c} a r e t h e v e r t i c e s o f Δ A B C \\ ∴ \vec{A B} = \vec{b} - \vec{a}, \vec{B C} = \vec{c} - \vec{b} a n d \vec{A c} = \vec{c} - \vec{a} \\ ∴ A r e a o f Δ A B C = \frac{1}{2} | \vec{A B} * \vec{A C} | \\ = \frac{1}{2} | (\vec{b} - \vec{a}) * (\vec{c} - \vec{b}) | = \frac{1}{2} | \vec{b} * \vec{c} - \vec{b} * \vec{a} - \vec{a} * \vec{c} + \vec{a} * \vec{a} | \\ = \frac{1}{2} | \vec{b} * \vec{c} + \vec{a} * \vec{b} + \vec{c} * \vec{a} | [\begin{array}{l} ? \vec{a} * \vec{b} = - \vec{b} * \vec{a} \\ \vec{c} * \vec{a} = - \vec{a} * \vec{c} \\ \vec{a} * \vec{a} = \vec{0} \end{array}] \\ F o r t h r e e v e c t o r s a r e c o l l i n e a r, a r e a o f Δ A B C = 0 \\ ∴ \frac{1}{2} | \vec{b} * \vec{c} + \vec{a} * \vec{b} + \vec{c} * \vec{a} | = 0 \\ | \vec{a} * \vec{b} + \vec{b} * \vec{c} + \vec{c} * \vec{a} | = 0 \\ w h i c h i s t h e c o n d i t i o n o f c o l l i n e a r i t y o f \vec{a}, \vec{b} a n d \vec{c} . \\ L e t \hat{n} b e t h e u n i t v e c t o r n o r m a l t o t h e p l a n e o f t h e Δ A B C \\ ∴ \hat{n} = \frac{\vec{A B} * \vec{A C}}{| \vec{A B} * \vec{A C} |} = \frac{\vec{a} * \vec{b} + \vec{b} * \vec{c} + \vec{c} * \vec{a}}{| \vec{a} * \vec{b} + \vec{b} * \vec{c} + \vec{c} * \vec{a} |} \end{array}$

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If $\vec{a}, \vec{b}, \vec{c}$  are non-zero and $\vec{b}$  and $\vec{c}$  are non-collinear. $\vec{a} + 5 \vec{b}$  is collinear with $\vec{c}$  and $\vec{b} + 6 \vec{c}$  is collinear with $\vec{a}$ . If $\vec{a} + α \vec{b} + β \vec{b} = 0$ , then find α + β.

Alok Kumar Singh

$\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$

⇒ $\vec{a} + 5 \vec{b}$ = $\vec{c}$ …(1)

$\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$

⇒ $\vec{b} + 6 \vec{c} = μ \vec{a}$ …(2)

From (1) and (2)

$\vec{b} + 6 \vec{c} = μ (λ \vec{c} - 5 \vec{b})$

-> $(1 + 5 μ) \vec{b} + (6 - λ μ) \vec{c} = 0$

$? \vec{b}$ and $\vec{c}$

$\int \frac{(s i n x - c o s x) s i n^{2} x}{t a n x (s i n^{3} x + c o s^{3} x)} d x$

$\int \frac{(s i n x - c o s x) s i n x c o s x}{s i n^{3} x + c o s^{3} x} d x$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3} \int \frac{d t}{t}$

$= \frac{l n t}{3} + c$

$= \frac{l n | s i n^{3} x + c o s^{3} x |}{3} + c$

If $f (x) = \frac{(2^{x} + 2^{- x}) (t a n x) \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)}}{{(7 x^{2} - 3 x + 1)}^{3}}$ , then f′(0) is equal to

Alok Kumar Singh

$f (x) = \frac{(2^{x} + 2^{- x}) t a n x \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)}}{{(7 x^{2} - 3 x + 1)}^{3}}$

$f (x) = (2^{x} + 2^{- x}) . t a n x . \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)} . {(7 x^{2} - 3 x + 1)}^{- 3}$

$f' (x) = (2^{x} + 2^{- x}) . s e c^{2} x . \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)} . {(7 x^{2} - 3 x + 1)}^{- 3} + t a n x . (Q (x))$

$f' (0) = 2.1 \sqrt[]{\frac{π}{4}} . 1$

$= \sqrt[]{π}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Ten 2025

Maths NCERT Exemplar Solutions Class 12th Chapter Ten 2025

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