b¯ and c¯ are non collinear vectors and c¯.c¯a¯=c¯If (a¯⋅c¯)b¯−(a¯⋅b¯)c¯+(a¯⋅b¯)b¯=(4−2β−sin⁡α)b¯+(β2−1)c¯ then Sin⁡(α+β)=

a→×(b→×c→)+(a→⋅b→)b→=(4−2β−sin⁡α)b→+β2−1c→ ⇒(a→⋅c→)b→−(a→⋅b→)c→+(a→⋅b→)b→=(4−2β−sin⁡α)b→+β2−1c→Comparing LHS and RHSa→⋅c→+a→⋅b→=(4−2β−sin⁡α)…(1)a→⋅b→=1−β2…(2)From (1) and (2)a→⋅c→=β2−2β+3−sin⁡α…(3)Now it is also given that(c→⋅c→)a→=c→⇒a→=c→c→⋅⋅0rca→⋅c→=c→⋅c→c→⋅c→1Therefore from eq (3)β2−2β+2−sin⁡α=0⇒sin⁡α=1+1−2β+β2=1+(1−β)2∴α=2nπ+π2, where n∈I

b¯ and c¯ are non collinear vectors and c¯.c¯a¯=c¯If (a¯⋅c¯)b¯−(a¯⋅b¯)c¯+(a¯⋅b¯)b¯=(4−2β−sin⁡α)b¯+(β2−1)c¯ then Sin⁡(α+β)=

a→×(b→×c→)+(a→⋅b→)b→=(4−2β−sin⁡α)b→+β2−1c→ ⇒(a→⋅c→)b→−(a→⋅b→)c→+(a→⋅b→)b→=(4−2β−sin⁡α)b→+β2−1c→Comparing LHS and RHSa→⋅c→+a→⋅b→=(4−2β−sin⁡α)…(1)a→⋅b→=1−β2…(2)From (1) and (2)a→⋅c→=β2−2β+3−sin⁡α…(3)Now it is also given that(c→⋅c→)a→=c→⇒a→=c→c→⋅⋅0rca→⋅c→=c→⋅c→c→⋅c→1Therefore from eq (3)β2−2β+2−sin⁡α=0⇒sin⁡α=1+1−2β+β2=1+(1−β)2∴α=2nπ+π2, where n∈I

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\bar{b}$ and $\bar{c}$ are non collinear vectors and $(\bar{c} . \bar{c}) \bar{a} = \bar{c}$
$If (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} + (\bar{a} \cdot \bar{b}) \bar{b} = (4 - 2 β - \sin α) \bar{b} + (β^{2} - 1) \bar{c} then Sin (α + β) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

cos 1

sin 1

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 β - \sin α) \vec{b} + (β^{2} - 1) \vec{c} \Rightarrow (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 β - \sin α) \vec{b} + (β^{2} - 1) \vec{c}$

Comparing LHS and RHS

$\begin{aligned} \vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{b} = (4 - 2 β - \sin α) \dots (1) \\ \vec{a} \cdot \vec{b} = 1 - β^{2} \dots (2) \end{aligned}$

From (1) and (2)

$\vec{a} \cdot \vec{c} = β^{2} - 2 β + 3 - \sin α \dots (3)$

Now it is also given that

$\begin{aligned} (\vec{c} \cdot \vec{c}) \vec{a} = \vec{c} \\ \Rightarrow \vec{a} = \frac{\vec{c}}{\vec{c} \cdot} \cdot \frac{0 r}{c} \vec{a} \cdot \vec{c} = \frac{\vec{c} \cdot \vec{c}}{\vec{c} \cdot \vec{c}} 1 \end{aligned}$