CameraIcon

SearchIcon

MyQuestionIcon

1

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

Angle between Two Lines

A unit vector...

Question

A unit vector $\to d$ is equally inclined at an angle $α$ with the vectors $\to a = cos θ^i + sin θ^j$ , $\to b = - sin θ^i + cos θ^j, \to c = \to k$ then $α =$

A

{cos}^{- 1} (\frac{1}{\sqrt{3}})

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

B

{cos}^{- 1} (\frac{1}{3})

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

C

\frac{π}{4}

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

D

\frac{π}{2}

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

Open in App

Solution

The correct option is A ${cos}^{- 1} (\frac{1}{\sqrt{3}})$
$∣ ∣ \to a ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣ = 1$
$\Rightarrow \to d . \to a = \to d . \to b = \to d . \to c = cos α, \to c = \to k$
$\Rightarrow \to d . (\to a - \to k) = 0,, \to d . (\to b - \to k) = 0$
$∴ \to d$ is along $(\to a - \to k) \times (\to b - \to k)$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k cos θ & sin θ & - 1 - sin θ & cos θ & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= (cos θ - sin θ)^i + (cos θ + sin θ)^j +^k$
$∴ \to d = \frac{(cos θ - sin θ)^i + (cos θ + sin θ)^j +^k}{\sqrt{{(cos θ - sin θ)}^{2} + {(cos θ + sin θ)}^{2} + 1}}$
Consider ${(cos θ - sin θ)}^{2} + {(cos θ + sin θ)}^{2} + 1$
$= 2 {cos}^{2} θ + 2 {sin}^{2} θ + 1$
$= 2 ({cos}^{2} θ + {sin}^{2} θ) + 1$
$= 2 + 1 = 3$
$∴ \to d = \frac{(cos θ - sin θ)^i + (cos θ + sin θ)^j +^k}{\sqrt{3}}$
$cos α = \to d .^k$
$= \frac{1}{\sqrt{3}} [(cos θ - sin θ)^i + (cos θ + sin θ)^j +^k] .^k$
$= \frac{1}{\sqrt{3}}$ since $^k .^k = 1$ and $^i .^k =^j .^k = 0$
$cos α = \frac{1}{\sqrt{3}}$
$∴ α = {cos}^{- 1} (\frac{1}{\sqrt{3}})$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A unit vector

d

is equally inclined at an angle

α

with the vectors

a = cos θ . i + sin θ . j, b = - sin θ . i + cos = θ . j

c = k

α

\to a, \to b, \to c

are unit vectors equally inclined to each other at an angle

\frac{π}{3}

[\to b + \to c, \to c + \to a, \to a + \to b] =

α = {cos}^{- 1} (\frac{3}{5}), β = {tan}^{- 1} (\frac{1}{3}),

0 < α, β < \frac{π}{2},

α - β

a

b

are two unit vectors inclined at an angle

\frac{π}{3}

{a \times (b + a \times b)} \cdot b

Let the angle between vectors $\to a$ and $\to b$ is $\frac{π}{6}$ , between vectors $\to b$ and $\to c$ is $\frac{π}{4}$ and between vectors $\to c$ and $\to a$ is $\frac{π}{3}$ . The angle, the vector $\to a$ makes with the plane containing vectors $\to b$ and $\to c$ , is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Angle between Two Lines

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Angle between Two Lines

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon