The vector a 1 - - a 2 - - a 3 k- and b 1 - -b 2 - -b 3 k- are perpendicular to each other if -

The correct option is B a _ 1 b _ 2 + b _ 2 a _ 1 + a _ 3 b _ 2 = 0 For perpendicular (a_1î+a_2ĵ+a_3k̂)(b_1î+b_2ĵ+b_3k̂)=0 a_1b_1+a_2b ...

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Byju's Answer

Standard XII

Physics

Vector Component

The vector ...

Question

The vector $a_{1}^i + a_{2}^j + a_{3}^k$ and $b_{1}^i + b_{2}^j + b_{3}^k$ are perpendicular to each other if :

\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}}

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a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = 1

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a_{1} b_{2} + b_{2} a_{1} + a_{3} b_{2} = 0

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Solution

The correct option is B $a_{1} b_{2} + b_{2} a_{1} + a_{3} b_{2} = 0$
For perpendicular
$(a_{1}^i + a_{2}^j + a_{3}^k) (b_{1}^i + b_{2}^j + b_{3}^k) = 0 a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = 0$

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Similar questions

Q. If

\to a = a_{1}^i + a_{2}^j - a_{3}^k

and

\to b = b_{1}^i + b_{2}^j - b_{3}^k

then find the value of

\to a \times \to b

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

Q. Let

\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

.If

∣ ∣ \to c ∣ ∣ = 1

and

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

then

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

For terms $a^{2} b^{2}, a^{3} b^{3}$ and $a^{3} b^{2}$ , $a^{3} b^{3}$ is known as ___

Match the object in Column A to the formulae in Column B. Choose the option with correct matches.

" $\begin{matrix} A : & B : 1. V o l u m e o f C o n e & 1. π r^{2} h 2. V o l u m e o f S p h e r e & 2. \frac{4}{3} π r^{3} 3. V o l u m e o f C y l i n d e r & 3. \frac{π r^{2} h}{3} \end{matrix}$

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The vector a1^i+a2^j+a3^k and b1^i+b2^j+b3^k are perpendicular to each other if :

The correct option is B a1b2+b2a1+a3b2=0For perpendicular(a1^i+a2^j+a3^k)(b1^i+b2^j+b3^k)=0a1b1+a2b2+a3b3=0

The vector $a_{1}^i + a_{2}^j + a_{3}^k$ and $b_{1}^i + b_{2}^j + b_{3}^k$ are perpendicular to each other if :

The correct option is B $a_{1} b_{2} + b_{2} a_{1} + a_{3} b_{2} = 0$
For perpendicular
$(a_{1}^i + a_{2}^j + a_{3}^k) (b_{1}^i + b_{2}^j + b_{3}^k) = 0 a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = 0$