Let a- and b- be two non-collinear unit vectors- If u- -a- - -a- b-b- and v- -a-b- then -v- is-u-u- - -u-a-u- - -u-b-u- - u-a- - b-

The correct options are A |u⃗| C |u⃗| + |u⃗.b⃗|Let θ be the angle between a⃗ and b⃗. Since, a⃗ and a⃗ and a⃗ are non collinear vectors, then θ≠ 0 and θ≠π. We ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Unit Vectors

Let a⃗ and b⃗...

Question

Let $\to a$ and $\to b$ be two non-collinear unit vectors. If $\to u = \to a - (\to a . \to b) \to b$ and $\to v = \to a \times \to b,$ then $| \to v |$ is

| \to u |

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

| \to u | + | \to u . \to a |

No worries! We‘ve got your back. Try BYJU‘S free classes today!

| \to u | + | \to u . \to b |

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

| \to u | + \to u . (\to a + \to b)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $| \to u |$
C $| \to u | + | \to u . \to b |$
Let $θ$ be the angle between $\to a$ and $\to b$ . Since, $\to a$ and $\to a$ and $\to a$ are non - collinear vectors, then $θ \neq 0$ and $θ \neq π .$
We have. $\to a . \to b = | \to a | | \to a | cos θ$
$= cos θ [because | \to a | = 1, | \to b | = 1, g i v e n]$
\text{Now}, $\to u = \to a - (\to a . \to b) \to b \Rightarrow | \to u | = | \to a - (\to a . \to b) \to b) | \Rightarrow | \to u |^{2} = | \to a - (\to a . \to b) \to b |^{2} \Rightarrow | \to u |^{2} = | \to a - cos θ \to b |^{2} \Rightarrow | \to u |^{2} = | \to a |^{2} + {cos}^{2} θ | \to b |^{2} - 2 cos θ (\to a . \to b) \Rightarrow | \to u |^{2} = 1 + c o s^{2} θ - 2 c o s^{2} θ \Rightarrow | \to u |^{2} = 1 - c o s^{2} θ \Rightarrow | \to u |^{2} = {sin}^{2} θ A l s o, \to v = \to a \times \to b [g i v e n] \Rightarrow | \to v |^{2} = | \to a \times \to b |^{2} \Rightarrow | \to v |^{2} = | \to a |^{2} | \to b |^{2} . s i n^{2} θ \Rightarrow | \to v |^{2} = {sin}^{2} θ ∴ | \to u |^{2} = | \to v |^{2} N o w, \to u . \to a = [\to a - (\to a . \to b) \to b] . \to a = \to a . \to a - (\to a . \to b) (\to b . \to a) = (\to a)^{2} - {cos}^{2} θ = 1 - {cos}^{2} θ = {sin}^{2} θ ∴ | \to u | + | \to \to u . \to a | = sin θ {sin}^{2} θ \neq | \to v |$
$\to u . \to b = [\to a - (\to a . \to b) \to b] . \to b = \to a . \to b - (\to a . \to b) (\to b . \to b) = \to a . \to b - \to a . \to b | \to b |^{2} = \to a . \to b - \to a . \to b = 0 ∴ | \to u | + | \to u . \to b | = | \to u | + 0 = | \to u | = | \to v | Also, \to u . (\to a + \to b) = \to u . \to a + \to u . \to b = \to u . \to a \Rightarrow | \to u | + \to u . (\to a + \to b) = | \to u | + \to u . \to a \neq | \to v |$

Suggest Corrections

Let →a and →b be two non-collinear unit vectors. If →u=→a−(→a.→b)→b and →v=→a×→b, then |→v| is

Let $\to a$ and $\to b$ be two non-collinear unit vectors. If $\to u = \to a - (\to a . \to b) \to b$ and $\to v = \to a \times \to b,$ then $| \to v |$ is