If {arrow}{a}, {arrow}{b}, {arrow}{c} be 3 non-coplanar vectors and ≤ft({arrow}{a}-λ{arrow}{b}) ·≤ft({arrow}{b}-2 {arrow}{c}) ×≤ft({arrow}{c}+2 {arrow}{a})=0 then λ=

Correct Answer is : −14-1/4−41

Correct Answer is : −14-1/4−41

Test Index

AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

Question : 33 of 160

Marks: +1, -0

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

\left(\overset{\rightarrow}{a}-\lambda\overset{\rightarrow}{b}\right) \cdot\left(\overset{\rightarrow}{b}-2 \overset{\rightarrow}{c}\right) \times\left(\overset{\rightarrow}{c}+2 \overset{\rightarrow}{a}\right)=0

\lambda=

Solution:

a, b, c

are there non-coplanar vectors

({a}-\lambda{b}) \cdot[({b}-2 {c}) \times({c}+2 {a})]=0

\Rightarrow({a}-\lambda{b}) \cdot\{ b \times c+2(b \times a)-2(c \times c)-4(c \times a) \}=0

\Rightarrow(a-\lambda b)-b \times c+2(b \times a)-4(c \times a)\}=0

[\because a \times a=0]

\Rightarrow a \cdot(b \times c)+2 a \cdot(b \times a)-4 a \cdot(c \times a)

-\lambda b \cdot(b \times c)-2 \lambda b \cdot(b \times a)+4 \lambda b \cdot(c \times a)=0

\Rightarrow[{a} {b} {c}]+2 {a} {b} {a}]-4[{a} {c} {a}]-\lambda[{b} {b} {c}]

-2 \lambda[{b} {b} {a}]+4 \lambda|{b} {c} {a}|=0

∴ We know that in scalar triple product, if two vectors are same in box then its value is 0.

\Rightarrow[{a} {b} {c}]+2(0)-4(0)-\lambda(0)+2 \lambda(0)+4 \lambda[{b} {c} {a}]=0

\Rightarrow[{a} {b} {c}]=-4 \lambda[{b} {c} {a}]

\Rightarrow 1=-4 \lambda

[\because [{a} {b} {c}]=[{b} c {a}]]

\Rightarrow\lambda=-\frac{1}{4}

Go to Question:

Prev Question

Next Question