If a -2 -3 k- and b -2 -k- c -4 -3 -2 k- and a - b is perpendicular to c- then -A- 1B- 3C- 4D- 2

The correct option is D 2a+λ b is perpendicular to c ⇒ (a+λ b). c =0 ⇒ a.c +λ b.c =0 λ = a.c/b.c= ( î+2ĵ+3k̂).(4î 3ĵ 2k̂)/(î 2ĵ+k̂).(4î 3ĵ 2k̂)= 4 6 6/4+6 2=1 ...

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Byju's Answer

Standard IX

Mathematics

Applications of Dot Product

If a =-î+2 ĵ+...

Question

If $a = -^i + 2^j + 3^k$ and $b =^i - 2^j +^k$ , $c = 4^i - 3^j - 2^k$ and $a + λ b$ is perpendicular to c, then $λ$ =

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Solution

The correct option is B 2
$a + λ b$ is perpendicular to $c \Rightarrow (a + λ b) . c = 0 \Rightarrow a . c + λ b . c = 0$
$λ = - \frac{a . c}{b . c} = - \frac{(-^i + 2^j + 3^k) . (4^i - 3^j - 2^k)}{(^i - 2^j +^k) . (4^i - 3^j - 2^k)} = - \frac{- 4 - 6 - 6}{4 + 6 - 2} = \frac{16}{8} = 2$

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Similar questions

Q. If

a = -^i + 2^j + 3^k

and

b =^i - 2^j +^k

c = 4^i - 3^j - 2^k

and

a + λ b

is perpendicular to c, then

λ

If $a = 2^i + 2^j + 3^k$ , $b = -^i + 2^j +^k$ and $c = 3^i +^j$ such that $a + λ b$ is perpendicular to c, then find the value of $λ$ .

Q. If

\to a = 2^i + 3^j - 4^k, \to b =^i +^j +^k

and

\to c = 4^i + 2^j + 3^k,

then

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

Q. If

\to a = 2^i + 2^j + 3^k,

\to b = -^i + 2^j +^k

and

\to c = 3^i +^j

are such that

\to a + λ \to b

is perpendicular to

\to c

then find the value of

λ

Q. If

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

and

\to c =^i + 3^j - (λ^{2} + 3 λ)^k

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λ

is a constant and

\to a

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If a=−^i+2^j+3^k and b=^i−2^j+^k, c=4^i−3^j−2^k and a+λb is perpendicular to c, then λ =

The correct option is B 2a+λb is perpendicular to c⇒(a+λb).c=0⇒a.c+λb.c=0 λ=−a.cb.c=−(−^i+2^j+3^k).(4^i−3^j−2^k)(^i−2^j+^k).(4^i−3^j−2^k)=−−4−6−64+6−2=168=2

If $a = -^i + 2^j + 3^k$ and $b =^i - 2^j +^k$ , $c = 4^i - 3^j - 2^k$ and $a + λ b$ is perpendicular to c, then $λ$ =

The correct option is B 2
$a + λ b$ is perpendicular to $c \Rightarrow (a + λ b) . c = 0 \Rightarrow a . c + λ b . c = 0$
$λ = - \frac{a . c}{b . c} = - \frac{(-^i + 2^j + 3^k) . (4^i - 3^j - 2^k)}{(^i - 2^j +^k) . (4^i - 3^j - 2^k)} = - \frac{- 4 - 6 - 6}{4 + 6 - 2} = \frac{16}{8} = 2$