MyQuestionIcon

1

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

Multiplication with Vectors

Find the unit...

Question

Find the unit vectors perpendicular to each of the following pairs of vectors:
$\to i + \to j + \to k$ and $\to i + \to 2 j - \to k$

A

a_{1} \to i + a_{2} \to j + a_{3} \to k,

where

a_{1} = \mp 3 / \sqrt{14}, a_{2} = \pm 2 / \sqrt{14}, a_{3} = \pm 1 / \sqrt{14}

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

B

a_{1} \to i + a_{2} \to j + a_{3} \to k,

where

a_{1} = \mp 3 / \sqrt{14}, a_{2} = \pm 2 / \sqrt{14}, a_{3} = \pm 2 / \sqrt{14}

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

C

a_{1} \to i + a_{2} \to j + a_{3} \to k,

where

a_{1} = \mp 2 / \sqrt{14}, a_{2} = \pm 2 / \sqrt{14}, a_{3} = \pm 1 / \sqrt{14}

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

D

a_{1} \to i + a_{2} \to j + a_{3} \to k,

where

a_{1} = \mp 3 / \sqrt{14}, a_{2} = \pm 2 / \sqrt{14}, a_{3} = \pm 3 / \sqrt{14}

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

Open in App

Solution

The correct option is A $a_{1} \to i + a_{2} \to j + a_{3} \to k,$ where $a_{1} = \mp 3 / \sqrt{14}, a_{2} = \pm 2 / \sqrt{14}, a_{3} = \pm 1 / \sqrt{14}$
The vector perpendicular to both vector is vector parallel to the unit vector of resultant of cross product of the vectors given.
The cross product of both products $= (^i +^j +^k) \times (^i + 2^j -^k)$
$= - 3^i + 2^j +^k$
So, the unit vector parallel to the resultant vector is $= \pm \frac{1}{\sqrt{14}} (- 3^i + 2^j +^k)$

Suggest Corrections

thumbs-up

0

Similar questions

\to a = 3 \to i + p \to j + 3 \to k

\to b = 2 \to i + 3 \to j + q \to k

are parallel vectors if

p = \frac{9}{2}

q = 2

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

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

are parallel, then

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

2 y - x = 8; 5 y - x = 14, y - 2 x = 1

(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})

(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})

\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

a_{1}, a_{2}, a_{3}, a_{4}

b

real numbers such that

b + 4 \sum K = 1 a_{K} = 8

b^{2} + 4 \sum K = 1 a_{K}^{2} = 16

. The possible integral value of '

b

K_{a 1}, K_{a 2}

K_{a 3}

H_{3} P O_{4}

10^{- 3}, 10^{- 8}

10^{- 12}

respectively. If

K_{w} (H_{2} O) = 10^{- 14}

, then which is correct?

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Basic Operations

PHYSICS

Watch in App

explore_more_icon

Explore more

Multiplication with Vectors

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon