If a- b- and c- are non-coplanar unit vectors equally inclined to one another at an acute angle - and a-b-b-c-p a-q b-t c- then A- p-1-1-2 cos-B- q -1-1-2 cos-C- p -2 cos-1-2 cos 2 -D- q-2 cos-1-2 cos-

The correct options are A p=1√(1+2cosθ) D q= 2cosθ√(1+2cosθ)a×b+b×c=pa+qb+tc Taking dot product with nbsp;a,b and nbsp;c simultaneously, p+qcosθ+tcosθ = [a ...

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Byju's Answer

Standard XII

Mathematics

Global Maxima

If a⃗, b⃗ and...

Question

If $\to a, \to b$ and $\to c$ are non-coplanar unit vectors equally inclined to one another at an acute angle $θ$ and $\to a \times \to b + \to b \times \to c = p \to a + q \to b + t \to c$ , then

p = \frac{1}{\sqrt{1 + 2 cos θ}}

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q = \frac{1}{\sqrt{1 + 2 cos θ}}

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p = \frac{- 2 cos θ}{\sqrt{1 + 2 cos 2 θ}}

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q = \frac{- 2 cos θ}{\sqrt{1 + 2 cos θ}}

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Solution

The correct options are
A $p = \frac{1}{\sqrt{1 + 2 cos θ}}$
D $q = \frac{- 2 cos θ}{\sqrt{1 + 2 cos θ}}$
$\to a \times \to b + \to b \times \to c = p \to a + q \to b + t \to c$

Taking dot product with $\to a$ , $\to b$ and $\to c$ simultaneously,
$p + q cos θ + t cos θ = [\to a$ $\to b$ $\to c] \dots (1)$
$p cos θ + q + t cos θ = 0 \dots (2)$
$p cos θ + q cos θ + t = [\to a$ $\to b$ $\to c] \dots (3)$

Subtracting $(1)$ and $(3)$
$p (1 - cos θ) - t (1 - cos θ) = 0 \Rightarrow p = t$

Subtracting $(1)$ and $(2)$
$p (1 - cos θ) + q (cos θ - 1) = [\to a \to b \to c] \Rightarrow q = p - \frac{[\to a \to b \to c]}{1 - cos θ}$
Putting in $(2)$
$2 p cos θ + p - \frac{[\to a \to b \to c]}{1 - cos θ} = 0$
$\Rightarrow p = \frac{[\to a \to b \to c]}{(1 - cos θ) (2 cos θ + 1)}$

$t = \frac{[\to a \to b \to c]}{(1 - cos θ) (2 cos θ + 1)}$

$q = \frac{- 2 [\to a \to b \to c] cos θ}{(1 - cos θ) (2 cos θ + 1)}$

Now,
$[\to a \to b \to c]^{2} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a . \to a & \to a . \to b & \to a . \to c \to b . \to a & \to b . \to b & \to b . \to c \to c . \to a & \to c . \to b & \to c . \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & cos θ & cos θ cos θ & 1 & cos θ cos θ & cos θ & 1 \end{matrix} ∣ ∣ ∣ ∣$

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$
$[\to a \to b \to c]^{2} = (1 + 2 cos θ) . ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 cos θ & 1 & cos θ cos θ & cos θ & 1 \end{matrix} ∣ ∣ ∣ ∣$

$\Rightarrow [\to a \to b \to c]^{2} = (1 + 2 cos θ) . (1 - cos θ)^{2}$
$\Rightarrow [\to a \to b \to c] = (\sqrt{1 + 2 cos θ}) . (1 - cos θ)$

We get
$p = \frac{1}{\sqrt{1 + 2 cos θ}}$
$q = \frac{- 2 cos θ}{\sqrt{1 + 2 cos θ}}$
$t = \frac{1}{\sqrt{1 + 2 cos θ}}$

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Similar questions

Q. Let

\to a, \to b

and

\to c

be non-zero vectors such that

(\to a \times \to b) \times \to c = \frac{1}{3} | \to b | | \to c | \to a

. If

θ

is the acute angle between the vectors

\to b

and

\to c

, then

sin θ

equals?

Q. Let

\to a, \to b

and

\to c

be three non-coplanar unit vectors such that the angle between each pair of vectors is

\frac{π}{3} .

\to a \times \to b + \to b \times \to c = p \to a + q \to b + r \to c,

where

p, q

and

r

are scalars, then the value of

\frac{p^{2} + 2 q^{2} + r^{2}}{q^{2}}

Q. Let

\to a, \to b

and

\to c

br non-coplanar unit vectors equally inclined to one another at an acute angle

θ

. Then

[\to a \to b \to c]

in terms of

θ

is equal to

Q. Let

\to a, \to b,

and

\to c

be three non-coplanar unit vectors such that the angle between every pair of them is

\frac{π}{3}

. If

\to a \times \to b + \to b \times \to c = p \to a + q \to b + r \to c

, where

p, q

and

r

are scalars, then the value of

\frac{p^{2} + 2 q^{2} + r^{2}}{q^{2}}

Q. Let

\to a, \to b

and

\to c

be non-zero vectors such that

(\to a \times \to b) \times \to c = \frac{1}{3} | \to b | | \to c | \to a

. If

θ

is the acute angle between the vectors

¯ ¯ ¯ b

and

¯ ¯ ¯ c

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If →a,→b and →c are non-coplanar unit vectors equally inclined to one another at an acute angle θ and →a×→b+→b×→c=p→a+q→b+t→c, then

If $\to a, \to b$ and $\to c$ are non-coplanar unit vectors equally inclined to one another at an acute angle $θ$ and $\to a \times \to b + \to b \times \to c = p \to a + q \to b + t \to c$ , then