The unit vector parallel to the resultant of the vectors A -4 -3 -6 k- andB -3 -8 k- isA- 1-73 -6 -2 k-B- 1-73 -6 -2 k-C- 1-493 -6 -2 k-D- 1-493 -6 -2 k-

The correct option is A 1/7(3î+6ĵ 2k̂) Resultant of vectors A⃗ and B⃗ R⃗=A⃗+B⃗=4î+3ĵ+6k̂ î+3ĵ 8k̂ R⃗=3î+6ĵ 2k̂ R̂=R⃗/|R⃗|=3î+6ĵ 2k̂/√(3^2+6^2 +( 2)^2)=3î+6ĵ 2k ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Addition and Subtraction in Unit Vector Notation

The unit vect...

Question

The unit vector parallel to the resultant of the vectors
$\to A = 4^i + 3^j + 6^k$ and
$\to B = -^i + 3^j - 8^k$ is

$\frac{1}{7} (3^i + 6^j - 2^k)$

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

$\frac{1}{7} (3^i + 6^j + 2^k)$

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

$\frac{1}{49} (3^i + 6^j - 2^k)$

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

$\frac{1}{49} (3^i - 6^j + 2^k)$

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

Open in App

Solution

The correct option is A
$\frac{1}{7} (3^i + 6^j - 2^k)$

Resultant of vectors $\to A$ and $\to B$
$\to R = \to A + \to B = 4^i + 3^j + 6^k -^i + 3^j - 8^k$
$\to R = 3^i + 6^j - 2^k$
$^R = \frac{\to R}{| \to R |} = \frac{3^i + 6^j - 2^k}{\sqrt{3^{2} + 6^{2} + (- 2)^{2}}} = \frac{3^i + 6^j - 2^k}{7}$

Suggest Corrections

307

Similar questions

Q. The vector

\to a = 1 / 7 (2^i + 3^j + 6^k)

\to b = 1 / 7 (3^i - 6^j + 2^k)

\to c = 1 / 7 (6^i + 2^j - 3^k)

form

Let $\to a = \frac{1}{7} (2^i + 3^j + 6^k), \to b = \frac{1}{7} (6^i + 2^j - 3^k), \to c = c_{1}^i + c_{2}^j + c - 3^k$ and matrix $A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \frac{6}{7} & \frac{2}{7} & - \frac{3}{7} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$
If $A A^{T} = I, t h e n \to c$ is equal to

Q. Show that each of the given three vectors is a unit vector:

\frac{1}{7} (2^i + 3^j + 6^k), \frac{1}{7} (3^i - 6^j + 2^k), \frac{1}{7} (6^i + 2^j - 3^k)

.
Also, show that they are mutually perpendicular to each other.

Q. Show that each of the three given vectors is a unit vectors. also Show that they are mutually

⊥

to each other.

\to a = \frac{1}{7} (2^i + 3^j + 6^k)

\to b = \frac{1}{7} (3^i - 6^j + 2^k)

\to c = \frac{1}{7} (6^i + 2^j - 3^k)

Q. A unit vector perpendicular to the plane of

a = 2^i - 6^j - 3^k, b = 4^i + 3^j -^k

Join BYJU'S Learning Program

The unit vector parallel to the resultant of the vectors →A=4^i+3^j+6^k and →B=−^i+3^j−8^k is

The correct option is A 17(3^i+6^j−2^k) Resultant of vectors →A and →B →R=→A+→B=4^i+3^j+6^k−^i+3^j−8^k →R=3^i+6^j−2^k ^R=→R|→R|=3^i+6^j−2^k√32+62+(−2)2=3^i+6^j−2^k7

The unit vector parallel to the resultant of the vectors
$\to A = 4^i + 3^j + 6^k$ and
$\to B = -^i + 3^j - 8^k$ is

The correct option is A
$\frac{1}{7} (3^i + 6^j - 2^k)$

Resultant of vectors $\to A$ and $\to B$
$\to R = \to A + \to B = 4^i + 3^j + 6^k -^i + 3^j - 8^k$
$\to R = 3^i + 6^j - 2^k$
$^R = \frac{\to R}{| \to R |} = \frac{3^i + 6^j - 2^k}{\sqrt{3^{2} + 6^{2} + (- 2)^{2}}} = \frac{3^i + 6^j - 2^k}{7}$