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Property 7

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Question

Let $\to a =^i +^j +^k$ , $\to b =^i$ and $\to c = c_{1}^i + c_{2}^j + c_{3}^k$ then if $c_{2} = - 1$ and $c_{3} = 1$ show that no value of $c_{1}$ can make vector $(a, b)$ and $\to c$ coplanar.

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Solution

$[\to a \to b \to c] = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 0 & 0 c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$
$= c_{2} - c_{3}$
Given: $c_{2} = - 1$ and $c_{2} = 1$
$\Rightarrow [\to a \to b \to c] = c_{2} - c_{3} = - 1 - 1 = - 2 \neq 0$
$∴$ No value of $c_{1}$ can make vector $(a, b, c)$ coplanar.

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\to a =^i +^j +^k

\to b =^i

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

c_{1} = 1

c_{2} = 2

c_{3}

which makes vector

(a, b)

\to c

\to a =^i + 2^j +^k, \to b =^i -^j +^k, \to c =^i +^j -^k .

A vector coplanar to

\to a

\to b

has a projection along

\to c

\frac{1}{\sqrt{3}}

, then the vector is

\to a =^i +^j +^k

\to b =^i -^j +^k

\to c =^i +^j -^k

\to a =^i -^j -^k

(\to a \times \to b) \times (\to c \times \to d)

is a vector orthogonal to both

\to a =^j -^k

\to c =^i -^j -^k

\to b

\to a \times \to b + \to c = \to 0

\to a \cdot \to b = 3

\to a =^i +^j +^k, \to b =^i +^j -^k

\to c =^i -^j +^k

\to d =^i -^j -^k

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

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