I - If the vectors a- 1-x-2 - b-x-3-4 are mutually perpendicular- then x-2II - If a-2-3k- b-2-k-c-3- and a-tb- is perpendicular to c- then t-5

The correct option is D Only II is trueTwo non zero #160;vectors are perpendicular if their dot product is zero.For the first case, #160; 1 × x + x × ...

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Byju's Answer

Standard XII

Mathematics

Dot Product

I : If the ve...

Question

I : If the vectors $\to a = (1, x, - 2)$ , $\to b = (x, 3, - 4)$ are mutually perpendicular, then $x = 2$
II : If $\to a =^i + 2^j + 3^k, \to b = -^i + 2^j +^k, \to c = 3^i +^j$ and $\to a + t \to b$ is perpendicular to $\to c$ then $t = 5$

Only I is true

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Only II is true

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Both I and II are true

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Neither l nor ll are true

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Solution

The correct option is D Only II is true
Two non- zero vectors are perpendicular if their dot product is zero.
For the first case, $1 \times x + x \times 3 + (- 2) \times (- 4) = 0; \Rightarrow x = - 2$
For the second case, $3 (1 - t) + 2 + 2 t = 0; \Rightarrow t = 5$
Hence option $B$ is correct.

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Similar questions

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I : If the vectors →a=(1,x,−2) , →b=(x,3,−4) are mutually perpendicular, then x=2II : If →a=^i+2^j+3^k,→b=−^i+2^j+^k,→c=3^i+^j and →a+t→b is perpendicular to →c then t=5

The correct option is D Only II is trueTwo non- zero vectors are perpendicular if their dot product is zero.For the first case, 1×x+x×3+(−2)×(−4)=0;⇒x=−2For the second case, 3(1−t)+2+2t=0;⇒t=5Hence option B is correct.

I : If the vectors $\to a = (1, x, - 2)$ , $\to b = (x, 3, - 4)$ are mutually perpendicular, then $x = 2$
II : If $\to a =^i + 2^j + 3^k, \to b = -^i + 2^j +^k, \to c = 3^i +^j$ and $\to a + t \to b$ is perpendicular to $\to c$ then $t = 5$

The correct option is D Only II is true
Two non- zero vectors are perpendicular if their dot product is zero.
For the first case, $1 \times x + x \times 3 + (- 2) \times (- 4) = 0; \Rightarrow x = - 2$
For the second case, $3 (1 - t) + 2 + 2 t = 0; \Rightarrow t = 5$
Hence option $B$ is correct.