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Standard XII

Physics

Multiplication with Vectors

ABC is a righ...

Question

ABC is a right-angled triangle in which $∠ B = 90^{\circ}$ and $B C = a$ . If n points $L_{1}, L_{2} . . . . . . ., L_{n}$ on AB is divided in n+1 equal parts and $L_{1} M_{1}, L_{2} M_{2} . . . . ., L_{n} M_{n}$ are line segments parallel to BC and $M_{1}, M_{2} . . . . . ., M_{n}$ are on AC then the sum of lengths of $L_{1} M_{1}, L_{2} M_{2}, . . . . . . L_{n} M_{n}$ is

\frac{a (n + 1)}{2}

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\frac{a (n - 1)}{2}

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\frac{a n}{2}

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None of these

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Solution

The correct option is C $\frac{a n}{2}$
$tan θ = \frac{A B}{B C} = \frac{A B}{a} B L_{1} = L_{1} L_{2} = L_{2} L_{3} = k (n + 1) k = A B tan θ = \frac{(n + 1) k}{a} \frac{a tan θ}{n + 1} = k k_{i} = \frac{(i) (a tan θ)}{n + 1} A l s o, tan θ = \frac{k_{i}}{x_{i}} = \frac{(i) (a tan θ)}{(n + 1) x_{i}} x_{i} = \frac{a_{i}}{(n + 1)}$

Length Segment $L_{i} M_{i} = a - \frac{a_{i}}{(n + 1)}$
$\sum L_{i} M_{i} = a n \sum i = 1 (1) - \frac{a}{n + 1} n \sum i = 1 i = a n - \frac{a}{2} \frac{n (n + 1)}{(n + 1)} = \frac{a n}{2}$

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Q. ABC is a right angle triangle in which

∠ B = 90^{0}

and BC=a. If n points

L_{1}

L_{2}

, ...,

L_{n}

on AB are such that AB divided in n+1 equal parts and

L_{1} M_{1}

L_{2} M_{2}

, ...,

L_{n} M_{n}

are line segments parallel to BC and

M_{1}

M_{2}

, ...,

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are on AC then the sum of the lengths of

L_{1} M_{1}

L_{2} M_{2}

, ...,

L_{n} M_{n}

A B C

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∠ B = 90^{\circ}

and

B C = a

. If

n

points

L_{1}, L_{2}, . . ., L_{n}

A B

are such that

A B

is divided in

n + 1

equal parts and

L_{1} M_{1}, L_{2} M_{2}, . . . . . . L_{n} M_{n}

are line segments parallel to

B C

and

M_{1}, M_{2}, . . ., M_{n}

are on

A C

then the sum of the lengths of

L_{1} M_{1}, L_{2} M_{2}, . . . . . . L_{n} M_{n}

A B C

is a right angled

△

in which

∠ B = 90^{\circ}

and

B C = a

. If

n

points

L_{1}, L_{2}, L_{3}, . . ., L_{n}

A B

are such that

A B

is divided into

(n + 1)

equal parts and

L_{1} M_{1}, L_{2} M_{2}, . . . L_{n} M_{n}

are line segments parallel to

B C

and points

M_{1}, M_{2}, M_{3}, . . ., M_{n}

are on

A C

, then the sum of the lengths of

L_{1} M_{1}, L_{2} M_{2}, . . . L_{n} M_{n}

is :

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is a right angle triangle at

B

and

B C = 10

. If

100

points

L_{1}, L_{2}, L_{3}, \dots, L_{100}

A B

are such that

A B

is divided into

101

equal parts and

L_{1} M_{1}, L_{2} M_{2}, \dots, L_{100} M_{100}

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and points

M_{1}, M_{2}, M_{3}, \dots M_{100}

are on

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, then the sum of

L_{1} M_{1}, L_{2} M_{2}, \dots, L_{100} M_{100}

A B C

is a right angle triangle at

B

and

B C = 10.

100

points

L_{1}, L_{2}, L_{3}, \dots, L_{100}

A B

are such that

A B

is divided into

101

equal parts and

L_{1} M_{1}, L_{2} M_{2}, \dots, L_{100} M_{100}

are line segments parallel to

B C

and points

M_{1}, M_{2}, M_{3}, \dots, M_{100}

are on

A C,

then the sum of

L_{1} M_{1}, L_{2} M_{2}, \dots, L_{100} M_{100}

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ABC is a right-angled triangle in which ∠B=90∘and BC=a. If n points L1,L2.......,Ln on AB is divided in n+1 equal parts and L1M1,L2M2.....,LnMn are line segments parallel to BC and M1,M2......,Mn are on AC then the sum of lengths of L1M1,L2M2,......LnMn is

The correct option is C an2tanθ=ABBC=ABaBL1=L1L2=L2L3=k(n+1)k=ABtanθ=(n+1)kaatanθn+1=kki=(i)(atanθ)n+1Also,tanθ=kixi=(i)(atanθ)(n+1)xixi=ai(n+1)Length Segment LiMi=a−ai(n+1)∑LiMi=an∑i=1(1)−an+1n∑i=1i=an−a2n(n+1)(n+1)=an2