Let a ^ and b ^ are two non-collinear unit vectors. If u → = a ^ - a ^ · b ^ b ^ and v → = a ^ × b ^ , then | v → | is equal to

We have, a ^ = b ^ = 1 Given, v → = a ^ × b ^ = a ^ b ^ sin θ n ^ = sin θ n ^ where, θ is the angle between a ^ and b ^ . ∴ v → = sin θ Now, u → = a ^ - a ^ · b ^ b ^ = a ^ - a ^ b ^ cos θ b ^ u → = a ^ - cos θ b ^ ∴ u → = a ^ 2 + - cos θ b ^ 2 - 2 a ^ - cos θ b ^ cos θ ⇒ u → = 1 + cos 2 θ - 2 cos 2 θ ⇒ u → = 1 - cos 2 θ = sin 2 θ ∴ u → = sin θ So, v → = u →

Mathematics - Vector Question with Solution | TestHub

MathematicsVectorGeneral definitions, Angle between vectors, Section formula, Geometrical resultsEasy2 minPYQ_2017

MathematicsEasysingle choice

Let $\hat{a}$ and $\hat{b}$ are two non-collinear unit vectors. If $\vec{u} = \hat{a} - (\hat{a} \cdot \hat{b}) \hat{b}$ and $\vec{v} = \hat{a} \times \hat{b}$ , then $| \vec{v} |$ is equal to

Options:

Answer:

Solution:

We have, $|\hat{a}| = |\hat{b}| = 1$

Given, $\vec{v} = \hat{a} \times \hat{b} = |\hat{a}| |\hat{b}| \sin θ \hat{n} = \sin θ \hat{n}$

where, $θ$ is the angle between $\hat{a}$ and $\hat{b}$ .

$∴ |\vec{v}| = \sin θ$

Now, $\vec{u} = \hat{a} - (\hat{a} \cdot \hat{b}) \hat{b} = \hat{a} - (|\hat{a}| |\hat{b}| \cos θ) \hat{b}$

$\vec{u} = \hat{a} - (\cos θ) \hat{b}$

$∴ |\vec{u}| = \sqrt{{|\hat{a}|}^{2} + {|- \cos θ |\hat{b}||}^{2} - 2 |\hat{a}| |- \cos θ \hat{b}| \cos θ}$

$\Rightarrow |\vec{u}| = \sqrt{1 + \cos^{2} θ - 2 \cos^{2} θ}$

$\Rightarrow |\vec{u}| = \sqrt{1 - \cos^{2} θ} = \sqrt{\sin^{2} θ}$

$∴ |\vec{u}| = \sin θ$

So, $|\vec{v}| = |\vec{u}|$

Stream:BITSATSubject:MathematicsTopic:VectorSubtopic:General definitions, Angle between vectors, Section formula, Geometrical results

⏱ 2mℹ️ Source: PYQ_2017

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