Let a- b and c be three unit vectors such that -a - b -2 - -a - c -2 - 8-Then -a - 2b -2 - -a - 2c -2 is equal to-

|a b|^2 + |a c|^2 = 8 (a b ) .( a b ) + ( a c )( a c ) = 8 a^2 +b^2 2a· b + a^2 +c^2 2a· c = 8 2a^2 +b^2 + c^2 2a· b 2a· c = 8 a· b+a· c = 2 N ...

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

Byju's Answer

Standard XII

Physics

Introduction

Let a, b and ...

Question

Let $\to a, \to b$ and $\to c$ be three unit vectors such that ${∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to a - \to c ∣ ∣}^{2} = 8.$ Then ${∣ ∣ ∣ \to a + 2 \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to a + 2 \to c ∣ ∣}^{2}$ is equal to:

Open in App

Solution

$| \to a - \to b |^{2} + | \to a - \to c |^{2} = 8$
$(\to a - \to b) . (\to a - \to b) + (\to a - \to c) (\to a - \to c) = 8$
$a^{2} + b^{2} - 2 a \cdot b + a^{2} + c^{2} - 2 a \cdot c = 8$

$2 a^{2} + b^{2} + c^{2} - 2 a \cdot b - 2 a \cdot c = 8$
$a \cdot b + a \cdot c = - 2$

Now $| \to a + 2 \to b |^{2} + | \to a + 2 \to c |^{2}$
$= 2 a^{2} + 4 b^{2} + 4 c^{2} + 4 a \cdot b + 4 a \cdot c$
$= 2 + 4 + 4 + 4 (- 2) = 2$

Suggest Corrections

Similar questions

Q. Let

\to a

and

\to b

be two unit vectors such that

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

. If

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

then

2 | \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. Let

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k,

and

c = c_{1} i + c_{2} j + c_{3} k

be three non-zero vectors such that

c

is unit vector perpendicular to both vectors

a

and

b

. If the angle between vectors

a

and

b

\frac{π}{6}

, then

{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

is equal to

Q. Let

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k

and

c = c_{1} i + c_{2} j + c_{3} k

be three non-zero such that

c

is a unit perpendicular to both vectors

a

and

b

. If the angle between vectors

a

and

b

\frac{π}{6},

then

{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

is equal to

Q. Let

\to a, \to b

and

\to c

be three vectors such that

\to a = \to b \times (\to b \times \to c)

. If magnitudes of the vectors

\to a, \to b

and

\to c

are

\sqrt{2}, 1

and

2

respectively and the angle between

\to b

and

\to c

θ (0 < θ < \frac{π}{2})

, then the value of

1 + tan θ

is equal to

Join BYJU'S Learning Program

Let →a,→b and →c be three unit vectors such that ∣∣∣→a−→b∣∣∣2+∣∣→a−→c∣∣2=8.Then ∣∣∣→a+2→b∣∣∣2+∣∣→a+2→c∣∣2 is equal to:

|→a−→b|2+|→a−→c|2=8 (→a−→b).(→a−→b)+(→a−→c)(→a−→c)=8 a2+b2−2a⋅b+a2+c2−2a⋅c=8 2a2+b2+c2−2a⋅b−2a⋅c=8 a⋅b+a⋅c=−2 Now |→a+2→b|2+|→a+2→c|2 =2a2+4b2+4c2+4a⋅b+4a⋅c =2+4+4+4(−2)=2

Let $\to a, \to b$ and $\to c$ be three unit vectors such that ${∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to a - \to c ∣ ∣}^{2} = 8.$ Then ${∣ ∣ ∣ \to a + 2 \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to a + 2 \to c ∣ ∣}^{2}$ is equal to: