Let veca, vecb and vecc be the three vectors having magnitudes 1,5 and 3 , respectively, such that the angle between veca and vecb is θ and veca ×( veca × vecb)= vecc. Then tan θ is equal to

Question

Let veca, vecb and vecc be the three vectors having magnitudes 1,5 and 3 , respectively, such that the angle between veca and vecb is θ and veca ×( veca × vecb)= vecc. Then tan θ is equal to (A) 0 (B) 2 / 3 (C) 3 / 5 (D) 3 / 4

Tardigrade Admin · Accepted Answer

veca ×( veca × vecb)= vecc ⇒| veca|| veca × vecb|=| vecc| (∵ veca perp( veca × vecb)) 1(1 × 5) sin θ=3 ⇒ sin θ=(3/5) ⇒ tan θ=(3/4) .

Q. Let $a, b$ and $c$ be the three vectors having magnitudes $1, 5$ and $3$ , respectively, such that the angle between $a$ and $b$ is $θ$ and $a \times (a \times b) = c$ . Then $tan θ$ is equal to

0

$2/3$

$3/5$

$3/4$

$a \times (a \times b) = c$
$\Rightarrow ∣ a ∣∣ a \times b ∣ = ∣ c ∣$
$(∵ a ⊥ (a \times b))$
$1 (1 \times 5) sin θ = 3$
$\Rightarrow sin θ = \frac{3}{5}$
$\Rightarrow tan θ = \frac{3}{4} .$

Q. Let a,b and c be the three vectors having magnitudes 1,5 and 3 , respectively, such that the angle between a and b is θ and a×(a×b)=c. Then tanθ is equal to

0

2/3

3/5

3/4

a×(a×b)=c ⇒∣a∣∣a×b∣=∣c∣ (∵a⊥(a×b)) 1(1×5)sinθ=3 ⇒sinθ=53​ ⇒tanθ=43​.

Q. Let $a, b$ and $c$ be the three vectors having magnitudes $1, 5$ and $3$ , respectively, such that the angle between $a$ and $b$ is $θ$ and $a \times (a \times b) = c$ . Then $tan θ$ is equal to

$2/3$

$3/5$

$3/4$

$a \times (a \times b) = c$
$\Rightarrow ∣ a ∣∣ a \times b ∣ = ∣ c ∣$
$(∵ a ⊥ (a \times b))$
$1 (1 \times 5) sin θ = 3$
$\Rightarrow sin θ = \frac{3}{5}$
$\Rightarrow tan θ = \frac{3}{4} .$