Let a =α hat i +3 hat j - hat k , b =3 hat i -β hat j +4 hat k and c = hat i +2 hat j -2 hat k where α, β ∈ R, be three vectors. If the projection of a on c is (10/3) and b × c =-6 hat i +10 hat j +7 hat k , then the value of α+β equal to :

Question

Let a =α hat i +3 hat j - hat k , b =3 hat i -β hat j +4 hat k and c = hat i +2 hat j -2 hat k where α, β ∈ R, be three vectors. If the projection of a on c is (10/3) and b × c =-6 hat i +10 hat j +7 hat k , then the value of α+β equal to : (A) 3 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

( a ⋅ c /| c |)=(10/3) ⇒ (α+6+2/√1+4+4)=(10/3) ⇒ α=2 and | beginarrayccc hati hat j hat k 3 -β 4 1 2 -2 endarray|=-6 hat i + hat j + hat k ⇒ 2 β-8=-6 ⇒ β=1 ⇒ α+β=3

Q. Let a=αi^+3j^​−k^,b=3i^−βj^​+4k^ and c=i^+2j^​−2k^ where α,β∈R, be three vectors. If the projection of a on c is 310​ and b×c=−6i^+10j^​+7k^, then the value of α+β equal to :

3

4

5

6

∣c∣a⋅c​=310​ ⇒1+4+4​α+6+2​=310​⇒α=2 and ∣∣​i^31​j^​−β2​k^4−2​∣∣​=−6i^+j^​+k^ ⇒2β−8=−6⇒β=1 ⇒α+β=3

$\frac{a \cdot c}{∣ c ∣} = \frac{10}{3}$
$\Rightarrow \frac{α + 6 + 2}{1 + 4 + 4} = \frac{10}{3} \Rightarrow α = 2$
and $∣ ∣ \hat{i} 31 \hat{j} - β 2 \hat{k} 4 - 2 ∣ ∣ = - 6 \hat{i} + \hat{j} + \hat{k}$
$\Rightarrow 2 β - 8 = - 6 \Rightarrow β = 1$
$\Rightarrow α + β = 3$