If a- b- and c- are unit vectors such that a-2 b-2 c-0- then -a-c- is equal to

A+2b +2c = 0 ⇒ nbsp;a+2c = 2b ⇒ (a+2c)· (a+2c) = ( 2b )· ( 2b ) ⇒ 1 + 4 nbsp;a· nbsp;c +4= 4 ⇒ nbsp;a· nbsp;c = 14 Let the angle between a and c be ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Cross Product of Two Vectors

If a⃗, b⃗, an...

Question

If $\to a, \to b,$ and $\to c$ are unit vectors such that $\to a + 2 \to b + 2 \to c = \to 0,$ then $| \to a \times \to c |$ is equal to :

\frac{\sqrt{15}}{4}

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

\frac{1}{4}

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

\frac{15}{16}

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

\frac{\sqrt{15}}{16}

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

Open in App

Solution

The correct option is A $\frac{\sqrt{15}}{4}$
$\to a + 2 \to b + 2 \to c = \to 0 \Rightarrow \to a + 2 \to c = - 2 \to b \Rightarrow (\to a + 2 \to c) \cdot (\to a + 2 \to c) = (- 2 \to b) \cdot (- 2 \to b) \Rightarrow 1 + 4 \to a \cdot \to c + 4 = 4 \Rightarrow \to a \cdot \to c = - \frac{1}{4}$
Let the angle between $\to a and \to c be θ$
$\Rightarrow cos θ = - \frac{1}{4} \Rightarrow sin θ = \pm \frac{\sqrt{15}}{4}$
Now,
$| \to a \times \to c | = | \to a | | \to c | sin θ = \frac{\sqrt{15}}{4}$

Suggest Corrections

Similar questions

Q. If

\to a, \to b

, and

\to c

are unit vectors such that

\to a + 2 \to b + 2 \to c = \to 0

, the

| \to a \times \to c |

is equal to :

Q. If

\to a, \to b

and

\to c

are unit vectors such that:

\to a + 2 \to b + 2 \to c = \to 0

, then

| \to a \times \to c |

is equal to :

Q. Let

\to a

and

\to b

be two unit vectors such that

| \to a + \to b | = \sqrt{3} .

\to c = \to a + 2 \to b + 3 (\to a \times \to b),

then

2 | \to c |

is equal to:

Q. If

\to a, \to b, \to c

are unit vectors such that

\to a + \to b + \to c = \to 0

and

(\to a, \to b) = \frac{π}{3}

, then

∣ ∣ \to a \times \to b ∣ ∣ + ∣ ∣ \to b \times \to c ∣ ∣ + | \to c \times \to a | =

Q. If

\to a, \to b, \to c

are unit vectors, then the value of

{∣ ∣ \to a - 2 \to b ∣ ∣}^{2} + {∣ ∣ \to b - 2 \to c ∣ ∣}^{2} + {| \to c - 2 \to a |}^{2}

does not exceed

Join BYJU'S Learning Program

If →a,→b, and →c are unit vectors such that →a+2→b+2→c=→0, then |→a×→c| is equal to :

The correct option is A √154→a+2→b+2→c=→0⇒→a+2→c=−2→b⇒(→a+2→c)⋅(→a+2→c)=(−2→b)⋅(−2→b)⇒1+4→a⋅→c+4=4⇒→a⋅→c=−14 Let the angle between →a and →c be θ ⇒cosθ=−14⇒sinθ=±√154 Now, |→a×→c|=|→a||→c|sinθ=√154

If $\to a, \to b,$ and $\to c$ are unit vectors such that $\to a + 2 \to b + 2 \to c = \to 0,$ then $| \to a \times \to c |$ is equal to :