Given four unit vectors a→,b→,c→ and d→. The vectors a→,b→ and c→ are coplanar but not collinear pair by pair and vector d→ is not coplanar with vectors a→,b→ and c→ and (a→ , b→)=(b→ , c→)=π3,(d→ , a→)=α and (d→, b→) =β, if(d→ , c→) =cos−1(mcosβ+ncosα) then |m−n| is: ( (x→,y→)=θ represents the angle between the vector between x→ and y→ is θ )

2Reason: Since a,b,ca,b,ca,b,c are unit and coplanar with (a,b)=(b,c)=π/3(a,b)=(b,c)=\pi/3(a,b)=(b,c)=π/3 and ddd is arbitrary, writec=m b+n ac= m\,b+n\,ac=mb+na.Using b⋅c=12b\cdot c=\tfrac12b⋅c=21 and ∣c∣2=1|c|^2=1∣c∣2=1 with a⋅b=12a\cdot b=\tfrac12a⋅b=21:m+n2=12,m2+n2+mn=1m+\frac{n}{2}=\frac12,\qquad m^2+n^2+mn=1m+2n=21,m2+n2+mn=1Solve to get n=±1n=\pm1n=±1 and m=1−n2m=\frac{1-n}{2}m=21−n. The pairwise non-collinearity rejects c=ac=ac=a (m=0,n=1)(m=0,n=1)(m=0,n=1), so m=1,n=−1m=1,n=-1m=1,n=−1 and c=b−ac=b-ac=b−a.Thuscos⁡(d,c)=d⋅c=mcos⁡β+ncos⁡α=cos⁡β−cos⁡α,\cos(d,c)=d\cdot c = m\cos\beta+n\cos\alpha=\cos\beta-\cos\alpha,cos(d,c)=d⋅c=mcosβ+ncosα=cosβ−cosα,so ∣m−n∣=∣1−(−1)∣=2.|m-n|=|1-(-1)|=2.∣m−n∣=∣1−(−1)∣=2..

Given four unit vectors a→,b→,c→ and d→. The vectors a→,b→ and c→ are coplanar but not collinear pair by pair and vector d→ is not coplanar with vectors a→,b→ and c→ and (a→ , b→)=(b→ , c→)=π3,(d→ , a→)=α and (d→, b→) =β, if(d→ , c→) =cos−1(mcosβ+ncosα) then |m−n| is: ( (x→,y→)=θ represents the angle between the vector between x→ and y→ is θ )

2Reason: Since a,b,ca,b,ca,b,c are unit and coplanar with (a,b)=(b,c)=π/3(a,b)=(b,c)=\pi/3(a,b)=(b,c)=π/3 and ddd is arbitrary, writec=m b+n ac= m\,b+n\,ac=mb+na.Using b⋅c=12b\cdot c=\tfrac12b⋅c=21 and ∣c∣2=1|c|^2=1∣c∣2=1 with a⋅b=12a\cdot b=\tfrac12a⋅b=21:m+n2=12,m2+n2+mn=1m+\frac{n}{2}=\frac12,\qquad m^2+n^2+mn=1m+2n=21,m2+n2+mn=1Solve to get n=±1n=\pm1n=±1 and m=1−n2m=\frac{1-n}{2}m=21−n. The pairwise non-collinearity rejects c=ac=ac=a (m=0,n=1)(m=0,n=1)(m=0,n=1), so m=1,n=−1m=1,n=-1m=1,n=−1 and c=b−ac=b-ac=b−a.Thuscos⁡(d,c)=d⋅c=mcos⁡β+ncos⁡α=cos⁡β−cos⁡α,\cos(d,c)=d\cdot c = m\cos\beta+n\cos\alpha=\cos\beta-\cos\alpha,cos(d,c)=d⋅c=mcosβ+ncosα=cosβ−cosα,so ∣m−n∣=∣1−(−1)∣=2.|m-n|=|1-(-1)|=2.∣m−n∣=∣1−(−1)∣=2..

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Given four unit vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d} .$ The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar but not collinear pair by pair and vector $\vec{d}$ is not coplanar with vectors $\vec{a}, \vec{b}$ and $\vec{c}$ and $(\vec{a}, \vec{b}) = (\vec{b}, \vec{c}) = \frac{π}{3}, (\vec{d}, \vec{a}) = α$ and $\overset{}{(\vec{d}, \vec{b})} = β, i f \overset{}{(\vec{d}, \vec{c})} = \cos^{- 1} (m \cos β + n \cos α)$ then $| m - n |$ is: ( $(\vec{x}, \vec{y}) = θ$ represents the angle between the vector between $\vec{x}$ and $\vec{y}$ is $θ$ )

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Detailed Solution

Reason: Since $a,b,c$ $a$ $,$ $b$ $,$ $c$ are unit and coplanar with $(a,b)=(b,c)=\pi/3$ $($ $a$ $,$ $b$ $)$ $=$ $($ $b$ $,$ $c$ $)$ $=$ $π$ $/3$ and $d$ $d$ is arbitrary, write
$c= m\,b+n\,a$ $c$ $=$ $mb$ $+$ $na$ .
Using $b\cdot c=\tfrac12$ $b$ $\cdot$ $c$ $=$ $21$ $$ and $|c|^2=1$ $∣$ $c$ $∣$ $2$ $=$ $1$ with $a\cdot b=\tfrac12$ $a$ $\cdot$ $b$ $=$ $21$ $$ :

m+\frac{n}{2}=\frac12,\qquad m^2+n^2+mn=1

$m$ $+$ $2$ $n$ $$ $=$ $21$ $$ $,$ $m$ $2$ $+$ $n$ $2$ $+$ $mn$ $=$ $1$

Solve to get $n=\pm1$ $n$ $=$ $\pm1$ and $m=\frac{1-n}{2}$ $m$ $=$ $21$ $-$ $n$ $$ . The pairwise non-collinearity rejects $c=a$ $c$ $=$ $a$ $(m=0,n=1)$ $($ $m$ $=$ $0$ $,$ $n$ $=$ $1$ $)$ , so $m=1,n=-1$ $m$ $=$ $1$ $,$ $n$ $=$ $-1$ and $c=b-a$ $c$ $=$ $b$ $-$ $a$ .