If x- y- z- are units vectors such that x-y-z-a-x-y- -z-b- a-x-3-2- a-y-7-4 and -a-2- then which of thefollowing is FALSE-A- y is perpendicular to zB- y is perpendicular to bC- Angle between x- and y- is acute-D- Angle between x and z is obtuse-

The correct option is B y nbsp;is perpendicular to bGiven :x+y+z=a |x+y+z|=|a| ⇒ |x+y+z|= 2 ⇒ nbsp;|x+y+z|^2=4 ⇒ (x+y+z)·(x+y+z) = 4 ⇒ 3+2(x·y+y·z+z·x) = 4 ...

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

If x⃗, y⃗, z⃗...

Question

If $\to x, \to y, \to z$ are units vectors such that $\to x + \to y + \to z = \to a$ , $(\to x \times \to y) \times \to z = \to b$ , $\to a \cdot \to x = \frac{3}{2}, \to a \cdot \to y = \frac{7}{4}$ and $| \to a | = 2,$ then which of the following is FALSE?

\to y

is perpendicular to

\to z

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\to y

is perpendicular to

\to b

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Angle between

\to x

and

\to y

is acute.

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Angle between

\to x

and

\to z

is obtuse.

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Solution

The correct option is B $\to y$ is perpendicular to $\to b$
Given : $\to x + \to y + \to z = \to a$
$| \to x + \to y + \to z | = | \to a | \Rightarrow | \to x + \to y + \to z | = 2 \Rightarrow | \to x + \to y + \to z |^{2} = 4 \Rightarrow (\to x + \to y + \to z) \cdot (\to x + \to y + \to z) = 4 \Rightarrow 3 + 2 (\to x \cdot \to y + \to y \cdot \to z + \to z \cdot \to x) = 4 \Rightarrow \to x \cdot \to y + \to y \cdot \to z + \to z \cdot \to x = \frac{1}{2} \dots (1)$

Now,
$\to x + \to y + \to z = \to a$
Taking dot product with $\to x$ and $\to y$ , we get
$1 + \to x \cdot \to y + \to x \cdot \to z = \frac{3}{2} \Rightarrow \to x \cdot \to y + \to x \cdot \to z = \frac{1}{2} \dots (2) \to x \cdot \to y + 1 + \to y \cdot \to z = \frac{7}{4} \Rightarrow \to x \cdot \to y + \to y \cdot \to z = \frac{3}{4} \dots (3)$
From equations $(1), (2)$ and $(3)$ , we get
$\to y \cdot \to z = 0 \to x \cdot \to z = - \frac{1}{4} \to x \cdot \to y = \frac{3}{4}$
Therefore,
$(\to x \times \to y) \times \to z = \to b \Rightarrow \to z \times (\to x \times \to y) = - \to b \Rightarrow (\to z \cdot \to y) \to x - (\to z \cdot \to x) \to y = - \to b \Rightarrow - \frac{1}{4} \to y = \to b$
Hence, $\to b$ is parallel to $\to y .$

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