If a - b - c - are three non-zero vectors- no two of which are collinear and the vector a - b - is collinear with c - b - c - is collinear with a - then a - b - c -a a -b b -c c -d none of these

(d) None of these a #8594;+b #8594; is collinear with c #8594; #8756; #8201;a #8594;+b #8594;=xc #8594; #160; #160; #160; #160; #16 ...

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Standard XII

Mathematics

Graph of Quadratic Expression

If a →, b →, ...

Question

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero vectors, no two of which are collinear and the vector $\vec{a} + \vec{b}$ is collinear with $\vec{c}, \vec{b} + \vec{c}$ is collinear with $\vec{a},$ then $\vec{a} + \vec{b} + \vec{c} =$
(a) $\vec{a}$

(b) $\vec{b}$

(c) $\vec{c}$

(d) none of these

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Solution

(d) None of these

$\vec{a} + \vec{b}$ is collinear with $\vec{c}$

$∴ \vec{a} + \vec{b} = x \vec{c} . . . . . (1)$

where x is scalar and x ≠ 0.

$\vec{b} + \vec{c}$ is collinear with $\vec{a}$

$\vec{b} + \vec{c} = y \vec{a} . . . . . (2)$

y is scalar and y ≠ 0

Substracting (2) from (1) we get,

$\begin{array}{l} \vec{a} - \vec{c} = x \vec{c} - y \vec{a} \\ \vec{a} (1 + y) = (1 + x) \vec{c} \end{array}$

As given $\vec{a}, \vec{c}$ are not collinear,

∴ 1 + y = 0 and 1 + x = 0

y = −1 and x = −1

Putting value of x in equation (1)

$\begin{array}{l} \vec{a} + \vec{b} = - \vec{c} \\ \vec{a} + \vec{b} + \vec{c} = 0 \end{array}$

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If a →, b →, c →are three non-zero vectors, no two of which are collinear and the vector a →+ b → is collinear with c →, b →+ c → is collinear with a →, then a →+ b →+ c →= (a) a → (b) b → (c) c → (d) none of these

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero vectors, no two of which are collinear and the vector $\vec{a} + \vec{b}$ is collinear with $\vec{c}, \vec{b} + \vec{c}$ is collinear with $\vec{a},$ then $\vec{a} + \vec{b} + \vec{c} =$
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(b) $\vec{b}$

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