Let a- and b- be two non-collinear unit vectors- If u - a - a - b b and v - a - b- then - v - isA- - u -B- - u - u - a -C- - u - u - b -D- - 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 ...

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Byju's Answer

Standard IX

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 |

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

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

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

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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. $\to a . \to b = | \to a | | \to a | c o s θ$
$= c o s θ [∵ | \to a | = 1, | \to b | = 1, g i v e n]$
Now, $\to u = \to a - (\to a . \to b) \to b \Rightarrow | \to u | = | \to a - (\to a . \to b) \to b) | \Rightarrow | \to u |^{2} = | \to a - (\to a . \to b) \to b |^{2} \Rightarrow | \to u |^{2} = | \to a - c o s θ \to b |^{2} \Rightarrow | \to u |^{2} = | \to a |^{2} + c o s^{2} θ | \to b |^{2} - 2 c o s θ (\to a . \to b) \Rightarrow | \to u |^{2} = 1 + c o s^{2} θ - 2 c o s^{2} θ \Rightarrow | \to u |^{2} = 1 - c o s^{2} θ \Rightarrow | \to u |^{2} = s i n^{2} θ A l s o, \to v = \to a \times \to b [g i v e n] \Rightarrow | \to v |^{2} = | \to a \times \to b |^{2} \Rightarrow | \to v |^{2} = | \to a |^{2} | \to b |^{2} . s i n^{2} θ \Rightarrow | \to v |^{2} = s i n^{2} θ ∴ | \to u |^{2} = | \to v |^{2} N o w, \to u . \to a = [\to a - (\to a . \to b) \to b] . \to a = \to a . \to a - (\to a . \to b) (\to b . \to a) = (\to a)^{2} - c o s^{2} θ = 1 - c o s^{2} θ = s i n^{2} θ ∴ | \to u | + | \to \to u . \to a | = s i n θ s i n^{2} θ \neq | \to v |$
$\to u . \to b = [\to a - (\to a . \to b) \to b] . \to b = \to a . \to b - (\to a . \to b) (\to b . \to b) = \to a . \to b - \to a . \to b | \to b |^{2} = \to a . \to b - \to a . \to b = 0 ∴ | \to u | + | \to u . \to b | = | \to u | + 0 = | \to u | = | \to v | A l s o, \to u . (\to a + \to b) = \to u . \to a + \to u . \to b = \to u . \to a \Rightarrow | \to u | + \to u . (\to a + \to b) = | \to u | + \to u . \to a \neq | \to v |$

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Similar questions

Q. Let

\to a

and

\to b

be two non-collinear vectors and

| \to a | = | \to b | = 1

. If

\to U = \to a - (\to a \cdot \to b) \to b

then

\sqrt{\to U \cdot \to a}

Q. 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 |

Q. Let

\to a

be a unit vector and

\to b

be a non-zero vector not parallel to

\to a

. If

\to u = \sqrt{3} (\to a \times \to b)

and

\to v = (\to a \times \to b) \times \to a

, then

(\to u, \to v) =

Q. Let

\to a

and

\to b

be two non collinear unit vectors. If

\to u = \to a - (\to a \cdot \to b) \to b

and

\to v = \to a \times \to b

then

∣ ∣ \to v ∣ ∣

Q. If

¯ ¯ ¯ u = ¯ ¯ ¯ a + ¯ ¯ b, ¯ ¯ ¯ v = ¯ ¯ ¯ a - ¯ ¯ b, | \to a | = | ¯ ¯ b | = 2

, then

| ¯ ¯ ¯ u \times ¯ ¯ ¯ v | =

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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