Let a-a1-2-a3k- b-b1-b2-b3k- and c-c1-c2-c3k- be three non-zero vectors such that c- is a unit vectorperpendicular to both the vectors a- and b- If the angle between a- and b- is -6- then begin vmatrix a1 -a2 -a3 11 b1 - b2 -b3 11 c1 -c2 -c3

The correct option is C 1/4 (a^2_1 + a^2_2 + a^2_3) (b^2_1 + b^2_2 + b^2_3) Since, (a⃗×b⃗) = |a⃗||b⃗| sin π/6. n̂ (a⃗×b⃗). c⃗ = 1/2 |a⃗||b⃗|.n̂.c⃗ [a⃗ b⃗ c⃗] ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Scalar Triple Product

Let a⃗=a1î+α2...

Question

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

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

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

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

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

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

Open in App

Solution

The correct option is C
$\frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$

Since, $(\to a \times \to b) = | \to a | | \to b | s i n \frac{π}{6} .^n (\to a \times \to b) . \to c = \frac{1}{2} | \to a | | \to b | .^n . \to c [\to a \to b \to c] = \frac{1}{2} | \to a | | \to b | . c o s 0^{\circ}$
$∵^n$ is perpendicular to both $\to a, \to b$ and $\to c$ is also a unit vector perpendicular to both $\to a$ and $\to b .$
$∴ {∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2} = {[\to a \to b \to c]}^{2} = \frac{1}{4} . | \to a |^{2} | \to b |^{2} = \frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$

Suggest Corrections

Similar questions

Q. Let the vectors

\to a, \to b, \to c

given as

a_{1}^i + a_{2}^j + a_{3}^k, b_{1}^i + b_{2}^j + b_{3}^k, c_{1}^i + c_{2}^j + c_{3}^k

. Then show that

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

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