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Vector Algebra Part 3

Section: Mathematics

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Question : 9 of 72

Marks: +1, -0

Let

\overset{\rightarrow}{a}, \overset{\rightarrow}{b}, \overset{\rightarrow}{c}

be three vectors such that

\overset{\rightarrow}{a}\times \overset{\rightarrow}{b}=2\left(\overset{\rightarrow}{a}\times \overset{\rightarrow}{c}\right)

\left|\overset{\rightarrow}{a}\right|=1,\left|\overset{\rightarrow}{b}\right|=4,\left|\overset{\rightarrow}{c}\right|=2

, and the angle between

\overset{\rightarrow}{b}

\overset{\rightarrow}{c}

60^{\circ}

\left|\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{c}\right|

[23 jan 2026 Shift 2]

2
0
1
4

Solution:

Concept:

If

\vec{a}\times(\vec{x})=0

, then

\vec{a}

and

\vec{x}

are parallel vectors.

Explanation:

Given

\vec{a}\times\vec{b}=2(\vec{a}\times\vec{c})

.

Rearranging:

\vec{a}\times\vec{b}-2(\vec{a}\times\vec{c})=\vec{a}\times(\vec{b}-2\vec{c})=0

.

Thus

\vec{b}-2\vec{c}

is parallel to

\vec{a}

, so

\vec{b}-2\vec{c}=\lambda\vec{a}

, i.e.,

\vec{b}=\lambda\vec{a}+2\vec{c}

.

Take square of magnitude:

|\vec{b}|^2 = \lambda^2|\vec{a}|^2 + 4|\vec{c}|^2 + 4\lambda(\vec{a}\cdot\vec{c})

.

Substitute

|\vec{a}|=1

,

|\vec{b}|=4

,

|\vec{c}|=2

:

16 = \lambda^2(1) + 4(4) + 4\lambda(\vec{a}\cdot\vec{c})

\Rightarrow \lambda^2 + 16 + 4\lambda(\vec{a}\cdot\vec{c}) = 16

\Rightarrow \lambda^2 + 4\lambda(\vec{a}\cdot\vec{c}) = 0

...(1)

Dot both sides of

\vec{b}=\lambda\vec{a}+2\vec{c}

with

\vec{c}

:

\vec{b}\cdot\vec{c} = \lambda(\vec{a}\cdot\vec{c}) + 2|\vec{c}|^2

.

Left:

|\vec{b}||\vec{c}|\cos60^\circ = 4\cdot2\cdot\frac12 = 4

.

Right:

\lambda(\vec{a}\cdot\vec{c}) + 2\cdot4 = \lambda(\vec{a}\cdot\vec{c}) + 8

.

Thus

4 = \lambda(\vec{a}\cdot\vec{c}) + 8

\Rightarrow \lambda(\vec{a}\cdot\vec{c}) = -4

...(2)

From (1) and (2):

\lambda^2 + 4(-4)=0

\Rightarrow \lambda^2=16

\Rightarrow \lambda = \pm 4

.

From (2):

\vec{a}\cdot\vec{c} = -4/\lambda

. For

\lambda=4

,

\vec{a}\cdot\vec{c} = -1

; for

\lambda=-4

,

\vec{a}\cdot\vec{c}=1

.

Hence

|\vec{a}\cdot\vec{c}| = 1

.

Answer:

Option C (1)

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