If a - 3k-b -k- and c - -2 -k-such that a-b-1 and b-c-3- then 13 a-b-c is equal to

A =αî+βĵ+ 3k̂ b = βî αĵ k̂ c = î 2 ĵ k̂ Now, a·b=1 ⇒ 2αβ 3=1 ⇒ nbsp;αβ= 2⋯(1) Also, b·c= 3 ⇒ nbsp; β +2α +1= 3 ⇒ nbsp;2α β= 4 Using equation (1), we get ...

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Standard XII

Mathematics

Equation of a Plane : Vector Form

If a =αî+βĵ+ ...

Question

If $\to a = α^i + β^j + 3^k,$ $\to b = - β^i - α^j -^k$ and
$\to c =^i - 2^j -^k$
such that $\to a \cdot \to b = 1$ and $\to b \cdot \to c = - 3$ , then $\frac{1}{3} ((\to a \times \to b) \cdot \to c)$ is equal to

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Solution

$\to a = α^i + β^j + 3^k \to b = - β^i - α^j -^k \to c =^i - 2^j -^k$

Now,
$\to a \cdot \to b = 1 \Rightarrow - 2 α β - 3 = 1 \Rightarrow α β = - 2 \dots (1)$
Also,
$\to b \cdot \to c = - 3 \Rightarrow - β + 2 α + 1 = - 3 \Rightarrow 2 α - β = - 4$
Using equation $(1)$ , we get
$2 α + \frac{2}{α} = - 4 \Rightarrow α^{2} + 2 α + 1 = 0 \Rightarrow α = - 1 \Rightarrow β = 2$
Therefore,
$\frac{1}{3} ((\to a \times \to b) \cdot \to c) = ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 2 & 3 - 2 & 1 & - 1 1 & - 2 & - 1 \end{matrix} ∣ ∣ ∣ ∣ = - 1 (- 1 - 2) - 2 (2 + 1) + 3 (4 - 1) = 3 - 6 + 9 = 6$

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If →a=α^i+β^j+3^k,→b=−β^i−α^j−^k and →c=^i−2^j−^k such that →a⋅→b=1 and →b⋅→c=−3, then 13((→a×→b)⋅→c) is equal to

If $\to a = α^i + β^j + 3^k,$ $\to b = - β^i - α^j -^k$ and
$\to c =^i - 2^j -^k$
such that $\to a \cdot \to b = 1$ and $\to b \cdot \to c = - 3$ , then $\frac{1}{3} ((\to a \times \to b) \cdot \to c)$ is equal to