If r - r - r -k- and - r -3- then r is equal toA- - 3-k-B- -3-k-C- -1-3-k-D- -1-3-k-

The correct option is B ±√(3)(î+ĵ+k̂)Let r=xî+yĵ+zk̂ nbsp; Since, nbsp;r . î= r . ĵ = r . k̂⇒ x = y =z Also |r| = √(x^2 + y^2 + z^2) = 3 ⇒ x = ±√(3) Hen ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If r ·î= r ·ĵ...

Question

If $\to r .^i = \to r .^j = \to r .^k and | \to r | = 3, then \to r$ is equal to

\pm 3 (^i +^j +^k)

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\pm \frac{1}{3} (^i +^j +^k)

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\pm \frac{1}{\sqrt{3}} (^i +^j +^k)

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\pm \sqrt{3} (^i +^j +^k)

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Solution

The correct option is D $\pm \sqrt{3} (^i +^j +^k)$
Let $\to r = x^i + y^j + z^k$
Since, $\to r .^i = \to r .^j = \to r .^k \Rightarrow x = y = z$
Also $| \to r | = \sqrt{x^{2} + y^{2} + z^{2}} = 3 \Rightarrow x = \pm \sqrt{3}$
Hence, the required vector is $\to r = \pm \sqrt{3} (^i +^j +^k)$

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Similar questions

Q. If

\to r .^i = \to r .^j = \to r .^k

and

| r | = 3

, then find

\to r

Q. If

\to r .^i = \to r .^j = \to r .^k

and

| r | = 2

, then find

\to r

Q. The unit vector perpendicular to both

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and

^j -^k

Q. The unit vector which, is perpendicular to the vectors

\to A =^i - 2^j +^k

and

\to B = 3^i - 2^j -^k

If $\to r \times \to b = \to c \times \to b$ and $\to r . \to a = 0$ where $\to a = 2^i + 3^j -^k, \to b = 3^i -^j +^k$ and $\to c =^i +^j + 3^k$ , then $\to r$ is equal to

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If →r.^i=→r.^j=→r.^k and |→r|=3, then →r is equal to

The correct option is D ±√3(^i+^j+^k)Let →r=x^i+y^j+z^k Since, →r.^i=→r.^j=→r.^k⇒x=y=z Also |→r|=√x2+y2+z2=3 ⇒x=±√3 Hence, the required vector is →r=±√3(^i+^j+^k)

If $\to r .^i = \to r .^j = \to r .^k and | \to r | = 3, then \to r$ is equal to

The correct option is D $\pm \sqrt{3} (^i +^j +^k)$
Let $\to r = x^i + y^j + z^k$
Since, $\to r .^i = \to r .^j = \to r .^k \Rightarrow x = y = z$
Also $| \to r | = \sqrt{x^{2} + y^{2} + z^{2}} = 3 \Rightarrow x = \pm \sqrt{3}$
Hence, the required vector is $\to r = \pm \sqrt{3} (^i +^j +^k)$