If a-b-a-c- and a-b-a-c- then

The correct option is C Either a⃗=0 #160;or #160; b⃗=c⃗ Given: #160; a⃗·b⃗ = a⃗·c⃗ ⇒a⃗· ( b⃗ c⃗ )=0 #160; #160;...(1) a⃗× ...

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Byju's Answer

Standard XII

Mathematics

Definition of Sets

If a⃗.b⃗=a⃗...

Question

If $\to a . \to b = \to a . \to c$ and $\to a \times \to b = \to a \times \to c$ , then

\to a

is perpendicular to

(\to b - \to c)

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Either

\to a = 0

\to b = \to c

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\to a

is parallel to

(\to b - \to c)

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\to a = \to b \neq \to c

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Solution

The correct option is C Either $\to a = 0$ or $\to b = \to c$
Given: $\to a \cdot \to b = \to a \cdot \to c$
$\Rightarrow \to a \cdot (\to b - \to c) = 0$ ...(1)
$\to a \times \to b = \to a \times \to c$
$\Rightarrow \to a \times (\to b - \to c) = 0$ ...(2)
So, angle between $\to a & (\to b - \to c)$ can't be $0^{\circ} & 90^{\circ}$ at some time.
So either $\to a = 0$ or $\to b - \to c = 0 \Rightarrow \to b = \to c$

Hence, option B.

Suggest Corrections

Similar questions

Q. Let

\to a, \to b, \to c

be vectors of length

3, 4, 5

respectively. Let

\to a

be perpendicular to

\to b + \to c, \to b

\to c + \to a

\to c

\to a + \to b .

Then

∣ ∣ \to a + \to b + \to c ∣ ∣

is:

Q. If

\to a, \to b

and

\to c

are perpendicular to

\to b + \to c, \to c + \to a

and

\to a + \to b

respectively and if

| \to a + \to b | = 6, | \to b + \to c | = 8

and

| \to c + \to a | = 10

, then

| \to a + \to b + \to c |

is equal to

Q. If

\to a, \to b, \to c

are vectors of lengths

2, 3, 4

. If

\to a

is orthogonal to

\to b + \to c, \to b

is orthogonal to

\to c + \to a

and

\to c

is orthogonal to

\to a + \to b

, then the modulus of

\to a + \to b + \to c

\to A, \to B

and

\to C

satisfy the relations,

\to A . \to B = 0

and

\to A . \to C = 0

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Q. If

| \to a | = 3, ∣ ∣ \to b ∣ ∣ = 4, | \to c | = 5, \to a ⊥ (\to b + \to c), \to b ⊥ (\to c + \to a)

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If →a.→b=→a.→c and →a×→b=→a×→c, then

The correct option is C Either →a=0 or →b=→cGiven: →a⋅→b=→a⋅→c⇒→a⋅(→b−→c)=0 ...(1)→a×→b=→a×→c⇒→a×(→b−→c)=0 ...(2)So, angle between →a & (→b−→c) can't be 0∘ & 90∘ at some time.So either →a=0 or →b−→c=0⇒→b=→cHence, option B.

If $\to a . \to b = \to a . \to c$ and $\to a \times \to b = \to a \times \to c$ , then