If p-k- and q-a-b-ck- where a-b-c-2-1-0-1-2- then the number of vectors q is perpendicular to a is

The correct option is C 19 p.q=(î+ĵ+k̂).(aî+bĵ+ck̂) =a+b+c ⇒î.î=ĵ.ĵ=k̂.k̂=1 and î.ĵ=ĵ.k̂=k̂.î=0 ⇒p.q=0 ⇒ a+b+c=0 ⇒ a=b ...

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Byju's Answer

Standard XII

Physics

Vector Component

If p=î+ĵ+k̂...

Question

If $\to p =^i +^j +^k$ and $\to q = a^i + b^j + c^k$ where $a, b, c \to {- 2, - 1, 0, 1, 2}$ . then the number of vectors $\to q$ is perpendicular to $\to a$ is

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Solution

The correct option is C $19$
$\to p . \to q = (^i +^j +^k) . (a^i + b^j + c^k)$
$= a + b + c$
$\Rightarrow^i .^i =^j .^j =^k .^k = 1$ and $^i .^j =^j .^k =^k .^i = 0$
$\Rightarrow \to p . \to q = 0$
$\Rightarrow a + b + c = 0$
$\Rightarrow a = b = c = 0$ ,
one vector $a = 0, b = 1, c = - 1, 6$ vectors by $a, b, c$
$a = 0, b = 2, c = - 2, 6$
vectors $a = b = 1, c = - 2, 3$
vectors $a = b = - 1, c = 2, 3$
vectors these are $1 + 6 + 6 + 3 + 3 = 19$ vectors

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If →p=^i+^j+^k and →q=a^i+b^j+c^k where a,b,c→{−2,−1,0,1,2}. then the number of vectors →q is perpendicular to →a is

The correct option is C 19→p.→q=(^i+^j+^k).(a^i+b^j+c^k)=a+b+c⇒^i.^i=^j.^j=^k.^k=1 and ^i.^j=^j.^k=^k.^i=0⇒→p.→q=0⇒a+b+c=0⇒a=b=c=0, one vectora=0,b=1,c=−1,6 vectors by a,b,ca=0,b=2,c=−2,6 vectors a=b=1,c=−2,3vectors a=b=−1,c=2,3 vectors these are 1+6+6+3+3=19 vectors

If $\to p =^i +^j +^k$ and $\to q = a^i + b^j + c^k$ where $a, b, c \to {- 2, - 1, 0, 1, 2}$ . then the number of vectors $\to q$ is perpendicular to $\to a$ is