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Question 22
Let r(t) =(2−3t) i+4tj−(1−t) k\mathbf { r } ( t ) = ( 2 - 3 t ) \mathbf { i } + 4 t \mathbf { j } - ( 1 - t ) \mathbf { k }r(t) =(2−3t) i+4tj−(1−t) k Then r′(0) \mathbf { r } ^ { \prime } ( 0 ) r′(0) is
A) −3i−4j−k- 3 \mathbf { i } - 4 \mathbf { j } - \mathbf { k }−3i−4j−k B) 3i+4j−k3 \mathbf { i } + 4 \mathbf { j } - \mathbf { k }3i+4j−k C) 3i+4j+k3 \mathbf { i } + 4 \mathbf { j } + \mathbf { k }3i+4j+k D) −3i−4j+k- 3 \mathbf { i } - 4 \mathbf { j } + \mathbf { k }−3i−4j+k E) −3i+4j+k- 3 \mathbf { i } + 4 \mathbf { j } + \mathbf { k }−3i+4j+k
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