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Test for Coplanarity

Let the vecto...

Question

Let the vectors $\to a, \to b, \to c$ be given as $a_{1}^i + a_{2}^j + a_{3}^k, b_{1}^i + b_{2}^j + b_{3}^k, c_{1}^i + c_{2}^j + c_{3}^k$ . Then show that $\to a \times (\to b + \to c) = \to a \times \to b + \to a \times \to c$

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Solution

$(¯ b + ¯ c)$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k a_{1} & a_{2} & a_{3} b_{1} + c_{1} & b_{2} + c_{2} & b_{3} + c_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$^i (a_{2} (b_{2} + c_{2}) - a_{2} (b_{2} + c_{2})) -^j$
$(a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1}) +^k$ ...... $(1)$
$¯ a \times ¯ b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (a_{2} b_{3} - a_{3} b_{2}) -^j (a_{1} b_{3} - a_{3} b_{1})$
$+^k (a_{1} b_{2} - a_{2} b_{1})$
$¯ a \times ¯ c = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k a_{1} & a_{2} & a_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (a_{2} c_{3} - a_{3} c_{2}) -^j (a_{1} c_{3} - a_{3} c_{1})$
$+^k (a_{1} c_{2} - a_{2} c_{1})$
$(¯ a \times ¯ b) + (¯ a \times ¯ c)$
$\Rightarrow^i (a_{2} b_{3} - a_{3} b_{2} + a_{2} c_{3} - a_{3} c_{2})$
$-^j (a_{1} b_{3} - a_{3} b_{1} + a_{1} c_{2} - a_{2} c_{1})$
$+^k (a_{1} b_{2} - a_{2} b_{1} + a_{1} c_{2} - a_{2} c_{1})$
$\Rightarrow^i (a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2}))$
$-^j (a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1}))$
$+^k (a_{1} (b_{2} + c_{2} - a_{2} (b_{1} + c_{1}))$ ....... $(2)$
from $(1)$ & $(2)$
$¯ a \times (¯ b + ¯ c) = (¯ a \times ¯ b) + (¯ a \times ¯ c)$

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Similar questions

\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k, \to c = c_{1}^i + c_{2}^j + c_{3}^k

∣ ∣ \to c ∣ ∣ = 1

(\to a \times \to b) \times \to c = 0

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

\to a = a_{1}^i + a_{2}^j + a_{3}^k; \to b = b_{1}^i + b_{2}^j + b_{3}^k;

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

[3 \to a + \to b, 3 \to b + \to c, 3 \to c + \to a]

= λ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a \cdot^i & \to a \cdot^j & \to a \cdot^k \to b \cdot^i & \to b \cdot^j & \to b \cdot^k \to c \cdot^i & \to c \cdot^j & \to c \cdot^k \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

then the value of

\frac{λ}{4}

\to a = a_{1}^i + a_{2}^j + a_{3}^k; \to b = b_{1}^i + b_{2}^j + b_{3}^k;

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

[3 \to a + \to b, 3 \to b + \to c, 3 \to c + \to a]

= λ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a \cdot^i & \to a \cdot^j & \to a \cdot^k \to b \cdot^i & \to b \cdot^j & \to b \cdot^k \to c \cdot^i & \to c \cdot^j & \to c \cdot^k \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

then the value of

\frac{λ}{4}

a_{1}^i + a_{2}^j + a_{3}^k

b_{1}^i + b_{2}^j + b_{3}^k

are perpendicular to each other if :

Let $\to a = a_{1}^i + α_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k a n d \to c = c_{1}^i + c_{2}^j + c_{3}^k$ be three non-zero vectors such that $\to c$ is a unit vector perpendicular to both the vectors $\to a$ and $\to b$ . If the angle between $\to a$ and $\to b$ is $\frac{π}{6},$
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

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