Let x- y- and z- be unit vectors such that x-y-z-a- x-y-z-b-x-y- -z-c-a - x -3-2- a - y -7-4 and - a -2Then which of the following options is-are CORRECT -A- a-z-3-2B- y-z-0C- z-4-3c-b-D- y -4 c

The correct options are B y·z=0 nbsp; C z=43(c b) nbsp;x+y+z=a Taking dot product with a, ⇒a·x+a·y+a·z=a·a=|a|^2=4 ⇒a·z=4 32 74=34 Now, taking dot product ...

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

Let x⃗, y⃗ an...

Question

Let $\to x, \to y$ and $\to z$ be unit vectors such that $\to x + \to y + \to z = \to a, \to x \times (\to y \times \to z) = \to b, (\to x \times \to y) \times \to z = \to c$ ,
$\to a \cdot \to x = \frac{3}{2}, \to a \cdot \to y = \frac{7}{4}$ and $| \to a | = 2$ .

Then which of the following option(s) is/are CORRECT ?

\to a \cdot \to z = \frac{3}{2}

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\to y \cdot \to z = 0

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\to z = \frac{4}{3} (\to c - \to b)

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\to y = 4 \to c

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Solution

The correct options are
B $\to y \cdot \to z = 0$
C $\to z = \frac{4}{3} (\to c - \to b)$

$\to x + \to y + \to z = \to a$
Taking dot product with $\to a$ ,
$\Rightarrow \to a \cdot \to x + \to a \cdot \to y + \to a \cdot \to z = \to a \cdot \to a = | \to a |^{2} = 4$
$\Rightarrow \to a \cdot \to z = 4 - \frac{3}{2} - \frac{7}{4} = \frac{3}{4}$

Now, taking dot product with $\to x$ and $\to y$ simultaneously,
$\to x \cdot (\to x + \to y + \to z) = \to x \cdot \to a = \to a \cdot \to x = \frac{3}{2}$
$\Rightarrow \to x \cdot \to y + \to x \cdot \to z = \frac{1}{2} \dots (1)$

Similarly, $\to y \cdot (\to x + \to y + \to z) = \to y \cdot \to a = \to a \cdot \to y = \frac{7}{4}$
$\Rightarrow \to x \cdot \to y + \to y \cdot \to z = \frac{3}{4} \dots (2)$

$| \to x + \to y + \to z |^{2} = | \to a |^{2}$
$\Rightarrow 3 + 2 (\to x \cdot \to y + \to y \cdot \to z + \to z \cdot \to x) = 4$
$\Rightarrow \to x \cdot \to y + \to y \cdot \to z + \to z \cdot \to x = \frac{1}{2} \dots (3)$

From $(1)$ and $(3)$ , we get
$\to y \cdot \to z = 0$
From $(2)$ and $(3)$ , we get
$\to x \cdot \to z = \frac{- 1}{4}$
From $(3)$ , we get
$\to x \cdot \to y = \frac{3}{4}$

Now, $\to x \times (\to y \times \to z) = \to b$
$\Rightarrow (\to x \cdot \to z) \to y - (\to x \cdot \to y) \to z = \to b$
$\Rightarrow \frac{- 1}{4} \to y - \frac{3}{4} \to z = \to b \dots (4)$

Also, $(\to x \times \to y) \times \to z = \to c$
$\Rightarrow \frac{- 1}{4} \to y = \to c$
$\Rightarrow \to y = - 4 \to c$
So, from $(4)$ , $\to z = \frac{4}{3} (\to c - \to b)$

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Let →x,→y and →z be unit vectors such that →x+→y+→z=→a, →x×(→y×→z)=→b, (→x×→y)×→z=→c, →a⋅→x=32, →a⋅→y=74 and |→a|=2. Then which of the following option(s) is/are CORRECT ?