If θ is the angle between the vectors 4 i ^ - k ^ and i ^ + j ^ + k ^ , then sin θ + cos θ equals to

Let a → = i ^ + j ^ + k ^ b → = 4 i ^ - k ^ Then, a → = 1 2 + 1 2 + 1 2 = 3 b → = 4 2 + 0 2 + - 4 2 = 4 2 Now, cos θ = a → · b → a → b → ⇒ cos θ = i ^ + j ^ + k ^ · 4 i ^ - 4 k ^ a → b → ⇒ cos θ = 4 - 4 a → b → = 0 Hence, sin θ = 1 - cos 2 θ = 1 Therefore, sin θ + cos θ = 1

Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Mathematics3D-Coordinate GeometryCoordinates of a point in space, Direction cosines,Easy2 minPYQ_2018

MathematicsEasysingle choice

If $θ$ is the angle between the vectors $4 (\hat{i} - \hat{k})$ and $\hat{i} + \hat{j} + \hat{k},$ then $(\sin θ + \cos θ)$ equals to

Options:

Answer:

Solution:

Let

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$

$\vec{b} = 4 (\hat{i} - \hat{k})$

Then,

$|\vec{a}| = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}$

$|\vec{b}| = \sqrt{4^{2} + 0^{2} + {(- 4)}^{2}} = 4 \sqrt{2}$

Now,

$\cos θ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

$\Rightarrow \cos θ = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot (4 \hat{i} - 4 \hat{k})}{|\vec{a}| |\vec{b}|}$

$\Rightarrow \cos θ = \frac{4 - 4}{|\vec{a}| |\vec{b}|} = 0$

Hence, $\sin θ = \sqrt{1 - \cos^{2} θ} = 1$

Therefore, $\sin θ + \cos θ = 1$

Stream:BITSATSubject:MathematicsTopic:3D-Coordinate GeometrySubtopic:Coordinates of a point in space, Direction cosines,

⏱ 2mℹ️ Source: PYQ_2018

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