MyQuestionIcon

1

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

Derivative from First Principle

Let a=a1î+a...

Question

Let $\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k$ and $\to c = c_{1}^i + c_{2}^j + c_{3}^k$ be three non-zero vectors such that $\to c$ is a unit vector perpendicular to both $\to a$ and $\to b$ and angle between $\to a$ and $\to b$ is $\frac{π}{6}$ , then
${⎡ ⎢ ⎣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎦}^{2} =$

A

0

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

B

1

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

C

\frac{1}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})

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

D

\frac{3}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2}) ({c_{1}}^{2} + {c_{2}}^{2} + {c_{3}}^{2})

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

Open in App

Solution

The correct option is D $\frac{1}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})$
${⎡ ⎢ ⎣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎦}^{2}$
$= {[\to a \to b \to c]}^{2}$
$= {[\to a \times \to b . \to c]}^{2}$
$= {∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2} {(\to n . \to c)}^{2}$
where $^n$ is unit vector along $\to a \times \to b, \to n = \to c$
$= {∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2} .1$
$= {∣ ∣ \to a ∣ ∣}^{2} {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} {sin}^{2} \frac{π}{6}$ (from the question)
$= \frac{1}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})$

Suggest Corrections

thumbs-up

0

Similar questions

\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k

\to c = c_{1}^i + c_{2}^j + c_{3}^k

\to c

is a unit vector orthogonal to the plane of

\to a, \to b

and the angle between

\to a

\to b

\frac{π}{3}

| \to c |^{2}

Let $\to a = a_{1}^i + α_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k a n d \to c = c_{1}^i + c_{2}^j + c_{3}^k$ be three non-zero vectors such that $\to c$ is a unit vector perpendicular to both the vectors $\to a$ and $\to b$ . If the angle between $\to a$ and $\to b$ is $\frac{π}{6},$
then ${∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}$ is equal to

\to a, \to b, \to c

be three non-zero vectors such that

\to c

is a unit vectors perpendicular is both

\to a

\to b

. If the angle between

\to a

\to b

\frac{π}{6}

[\to a \to b \to c]^{2} = \frac{1}{4} | \to a |^{2} | \to b |^{2}

Let the vectors and be such that and, then is a unit vector, if the angle between and is

(A) (B) (C) (D)

\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k, \to c = c_{1}^i + c_{2}^j + c_{3}^k

∣ ∣ \to c ∣ ∣ = 1

(\to a \times \to b) \times \to c = 0

{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2} =

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Derivative of Simple Functions

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Derivative from First Principle

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon