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Intercept Made by Circle on Axes

a,b,c are thr...

Question

$a, b, c$ are three non-coplanar vectors. Then the equation $\to r = (1 - p - q) \to a + p \to b + q \to c$ represents :

A

A straight line

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B

A plane

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C

A plane passing through the origin

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D

A sphere

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Solution

The correct option is C A plane
Let $\to r$ be also represented in the form of three non-coplanar vectors $\to a, \to b, \to c$ .
Let $\to r = x \to a + y \to b + z \to c$
Hence, $x \to a + y \to b + z \to c = (1 - p - q) \to a + p \to b + q \to c$
Comparing RHS to LHS gives us
$x = 1 - p - q$
$y = p$
$z = q$
Then $x = 1 - y - z$
$x + y + z = 1$
This represents an equation of a plane where the co-ordinate axes are $\to a$ , $\to b$ and $\to c$ , and also the plane does not pass through origin.

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Similar questions

¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

are three non-coplanar vectors,

¯ ¯ ¯ p, ¯ ¯ ¯ q, ¯ ¯ ¯ r

are non-zero vectors such that

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to q = \frac{\to c \times \to a}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

, then the value of the expression

(\to a + \to b + \to c) . (\to p + \to q + \to r)

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are three non-coplanar non-zero vectors and

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(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b) =

Q. Given three non-zero, non-coplanar vectors

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Q. Given, three non - zero, non -coplanar vectors

\to a

\to b

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_{1} = p \to a + q \to b + \to c

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